from collections import defaultdict
import random
from operator import gt

from sympy.core import Basic, C
from sympy.core.compatibility import is_sequence, iterable # logical location
from sympy.core.compatibility import as_int, quick_sort,\
    product as cartes, combinations, combinations_with_replacement
from sympy.utilities.misc import default_sort_key
from sympy.utilities.exceptions import SymPyDeprecationWarning

def flatten(iterable, levels=None, cls=None):
    """
    Recursively denest iterable containers.

    >>> from sympy.utilities.iterables import flatten

    >>> flatten([1, 2, 3])
    [1, 2, 3]
    >>> flatten([1, 2, [3]])
    [1, 2, 3]
    >>> flatten([1, [2, 3], [4, 5]])
    [1, 2, 3, 4, 5]
    >>> flatten([1.0, 2, (1, None)])
    [1.0, 2, 1, None]

    If you want to denest only a specified number of levels of
    nested containers, then set ``levels`` flag to the desired
    number of levels::

    >>> ls = [[(-2, -1), (1, 2)], [(0, 0)]]

    >>> flatten(ls, levels=1)
    [(-2, -1), (1, 2), (0, 0)]

    If cls argument is specified, it will only flatten instances of that
    class, for example:

    >>> from sympy.core import Basic
    >>> class MyOp(Basic):
    ...     pass
    ...
    >>> flatten([MyOp(1, MyOp(2, 3))], cls=MyOp)
    [1, 2, 3]

    adapted from http://kogs-www.informatik.uni-hamburg.de/~meine/python_tricks
    """
    if levels is not None:
        if not levels:
            return iterable
        elif levels > 0:
            levels -= 1
        else:
            raise ValueError("expected non-negative number of levels, got %s" % levels)

    if cls is None:
        reducible = lambda x: is_sequence(x, set)
    else:
        reducible = lambda x: isinstance(x, cls)

    result = []

    for el in iterable:
        if reducible(el):
            if hasattr(el, 'args'):
                el = el.args
            result.extend(flatten(el, levels=levels, cls=cls))
        else:
            result.append(el)

    return result

def unflatten(iter, n=2):
    """Group ``iter`` into tuples of length ``n``. Raise an error if
    the length of ``iter`` is not a multiple of ``n``.
    """
    if n < 1 or len(iter) % n:
        raise ValueError('iter length is not a multiple of %i' % n)
    return list(zip(*(iter[i::n] for i in range(n))))

def reshape(seq, how):
    """Reshape the sequence according to the template in ``how``.

    Examples
    ========

    >>> from sympy.utilities import reshape
    >>> seq = list(range(1, 9))

    >>> reshape(seq, [4]) # lists of 4
    [[1, 2, 3, 4], [5, 6, 7, 8]]

    >>> reshape(seq, (4,)) # tuples of 4
    [(1, 2, 3, 4), (5, 6, 7, 8)]

    >>> reshape(seq, (2, 2)) # tuples of 4
    [(1, 2, 3, 4), (5, 6, 7, 8)]

    >>> reshape(seq, (2, [2])) # (i, i, [i, i])
    [(1, 2, [3, 4]), (5, 6, [7, 8])]

    >>> reshape(seq, ((2,), [2])) # etc....
    [((1, 2), [3, 4]), ((5, 6), [7, 8])]

    >>> reshape(seq, (1, [2], 1))
    [(1, [2, 3], 4), (5, [6, 7], 8)]

    >>> reshape(tuple(seq), ([[1], 1, (2,)],))
    (([[1], 2, (3, 4)],), ([[5], 6, (7, 8)],))

    >>> reshape(tuple(seq), ([1], 1, (2,)))
    (([1], 2, (3, 4)), ([5], 6, (7, 8)))
    """
    m = sum(flatten(how))
    n, rem = divmod(len(seq), m)
    if m < 0 or rem:
        raise ValueError('template must sum to positive number '
        'that divides the length of the sequence')
    i = 0
    container = type(how)
    rv = [None]*n
    for k in range(len(rv)):
        rv[k] = []
        for hi in how:
            if type(hi) is int:
                rv[k].extend(seq[i: i + hi])
                i += hi
            else:
                n = sum(flatten(hi))
                fg = type(hi)
                rv[k].append(fg(reshape(seq[i: i + n], hi))[0])
                i += n
        rv[k] = container(rv[k])
    return type(seq)(rv)

def group(container, multiple=True):
    """
    Splits a container into a list of lists of equal, adjacent elements.

    >>> from sympy.utilities.iterables import group

    >>> group([1, 1, 1, 2, 2, 3])
    [[1, 1, 1], [2, 2], [3]]

    >>> group([1, 1, 1, 2, 2, 3], multiple=False)
    [(1, 3), (2, 2), (3, 1)]

    """
    if not container:
        return []

    current, groups = [container[0]], []

    for elem in container[1:]:
        if elem == current[-1]:
            current.append(elem)
        else:
            groups.append(current)
            current = [elem]

    groups.append(current)

    if multiple:
        return groups

    for i, current in enumerate(groups):
        groups[i] = (current[0], len(current))

    return groups

def postorder_traversal(node, key=None):
    """
    Do a postorder traversal of a tree.

    This generator recursively yields nodes that it has visited in a postorder
    fashion. That is, it descends through the tree depth-first to yield all of
    a node's children's postorder traversal before yielding the node itself.

    Parameters
    ==========
    node : sympy expression
        The expression to traverse.
    key : (default None) sort key
        The key used to sort args of Basic objects. When None, args of Basic
        objects are processed in arbitrary order.

    Returns
    =======
    subtree : sympy expression
        All of the subtrees in the tree.

    Examples
    ========
    >>> from sympy import symbols, default_sort_key
    >>> from sympy.utilities.iterables import postorder_traversal
    >>> from sympy.abc import w, x, y, z

    The nodes are returned in the order that they are encountered unless key
    is given.

    >>> list(postorder_traversal(w + (x + y)*z)) # doctest: +SKIP
    [z, y, x, x + y, z*(x + y), w, w + z*(x + y)]
    >>> list(postorder_traversal(w + (x + y)*z, key=default_sort_key))
    [w, z, x, y, x + y, z*(x + y), w + z*(x + y)]


    """
    if isinstance(node, Basic):
        args = node.args
        if key:
            args = list(args)
            args.sort(key=key)
        for arg in args:
            for subtree in postorder_traversal(arg, key):
                yield subtree
    elif iterable(node):
        for item in node:
            for subtree in postorder_traversal(item, key):
                yield subtree
    yield node

def interactive_traversal(expr):
    """Traverse a tree asking a user which branch to choose. """
    from sympy.printing import pprint

    RED, BRED = '\033[0;31m', '\033[1;31m'
    GREEN, BGREEN = '\033[0;32m', '\033[1;32m'
    YELLOW, BYELLOW = '\033[0;33m', '\033[1;33m'
    BLUE, BBLUE = '\033[0;34m', '\033[1;34m'
    MAGENTA, BMAGENTA = '\033[0;35m', '\033[1;35m'
    CYAN, BCYAN = '\033[0;36m', '\033[1;36m'
    END = '\033[0m'

    def cprint(*args):
        print("".join(map(str, args)) + END)

    def _interactive_traversal(expr, stage):
        if stage > 0:
            print()

        cprint("Current expression (stage ", BYELLOW, stage, END, "):")
        print(BCYAN)
        pprint(expr)
        print(END)

        if isinstance(expr, Basic):
            if expr.is_Add:
                args = expr.as_ordered_terms()
            elif expr.is_Mul:
                args = expr.as_ordered_factors()
            else:
                args = expr.args
        elif hasattr(expr, "__iter__"):
            args = list(expr)
        else:
            return expr

        n_args = len(args)

        if not n_args:
            return expr

        for i, arg in enumerate(args):
            cprint(GREEN, "[", BGREEN, i, GREEN, "] ", BLUE, type(arg), END)
            pprint(arg)
            print()

        if n_args == 1:
            choices = '0'
        else:
            choices = '0-%d' % (n_args-1)

        try:
            choice = input("Your choice [%s,f,l,r,d,?]: " % choices)
        except EOFError:
            result = expr
            print()
        else:
            if choice == '?':
                cprint(RED, "%s - select subexpression with the given index" % choices)
                cprint(RED, "f - select the first subexpression")
                cprint(RED, "l - select the last subexpression")
                cprint(RED, "r - select a random subexpression")
                cprint(RED, "d - done\n")

                result = _interactive_traversal(expr, stage)
            elif choice in ['d', '']:
                result = expr
            elif choice == 'f':
                result = _interactive_traversal(args[0], stage+1)
            elif choice == 'l':
                result = _interactive_traversal(args[-1], stage+1)
            elif choice == 'r':
                result = _interactive_traversal(random.choice(args), stage+1)
            else:
                try:
                    choice = int(choice)
                except ValueError:
                    cprint(BRED, "Choice must be a number in %s range\n" % choices)
                    result = _interactive_traversal(expr, stage)
                else:
                    if choice < 0 or choice >= n_args:
                        cprint(BRED, "Choice must be in %s range\n" % choices)
                        result = _interactive_traversal(expr, stage)
                    else:
                        result = _interactive_traversal(args[choice], stage+1)

        return result

    return _interactive_traversal(expr, 0)

def variations(seq, n, repetition=False):
    """Returns a generator of the n-sized variations of ``seq`` (size N).
    ``repetition`` controls whether items in ``seq`` can appear more than once;

    Examples
    ========

    variations(seq, n) will return N! / (N - n)! permutations without
    repetition of seq's elements:

        >>> from sympy.utilities.iterables import variations
        >>> list(variations([1, 2], 2))
        [(1, 2), (2, 1)]

    variations(seq, n, True) will return the N**n permutations obtained
    by allowing repetition of elements:

        >>> list(variations([1, 2], 2, repetition=True))
        [(1, 1), (1, 2), (2, 1), (2, 2)]

    If you ask for more items than are in the set you get the empty set unless
    you allow repetitions:

        >>> list(variations([0, 1], 3, repetition=False))
        []
        >>> list(variations([0, 1], 3, repetition=True))[:4]
        [(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1)]

    See Also
    ========

    sympy.core.compatibility.permutations
    """
    from sympy.core.compatibility import permutations

    if not repetition:
        seq = tuple(seq)
        if len(seq) < n:
            return
        for i in permutations(seq, n):
            yield i
    else:
        if n == 0:
            yield ()
        else:
            for i in range(len(seq)):
                for cc in variations(seq, n - 1, True):
                    yield (seq[i],) + cc

def subsets(seq, k=None, repetition=False):
    """Generates all k-subsets (combinations) from an n-element set, seq.

       A k-subset of an n-element set is any subset of length exactly k. The
       number of k-subsets of an n-element set is given by binomial(n, k),
       whereas there are 2**n subsets all together. If k is None then all
       2**n subsets will be returned from shortest to longest.

       Examples
       ========
       >>> from sympy.utilities.iterables import subsets

       subsets(seq, k) will return the n!/k!/(n - k)! k-subsets (combinations)
       without repetition, i.e. once an item has been removed, it can no
       longer be "taken":

           >>> list(subsets([1, 2], 2))
           [(1, 2)]
           >>> list(subsets([1, 2]))
           [(), (1,), (2,), (1, 2)]
           >>> list(subsets([1, 2, 3], 2))
           [(1, 2), (1, 3), (2, 3)]


       subsets(seq, k, repetition=True) will return the (n - 1 + k)!/k!/(n - 1)!
       combinations *with* repetition:

           >>> list(subsets([1, 2], 2, repetition=True))
           [(1, 1), (1, 2), (2, 2)]

       If you ask for more items than are in the set you get the empty set unless
       you allow repetitions:

           >>> list(subsets([0, 1], 3, repetition=False))
           []
           >>> list(subsets([0, 1], 3, repetition=True))
           [(0, 0, 0), (0, 0, 1), (0, 1, 1), (1, 1, 1)]

       """
    if k is None:
        for k in range(len(seq) + 1):
            for i in subsets(seq, k, repetition):
                yield i
    else:
        if not repetition:
            for i in combinations(seq, k):
                yield i
        else:
            for i in combinations_with_replacement(seq, k):
                yield i

def numbered_symbols(prefix='x', cls=None, start=0, *args, **assumptions):
    """
    Generate an infinite stream of Symbols consisting of a prefix and
    increasing subscripts.

    Parameters
    ----------
    prefix : str, optional
        The prefix to use. By default, this function will generate symbols of
        the form "x0", "x1", etc.

    cls : class, optional
        The class to use. By default, it uses Symbol, but you can also use Wild.

    start : int, optional
        The start number.  By default, it is 0.

    Returns
    -------
    sym : Symbol
        The subscripted symbols.
    """
    if 'dummy' in assumptions:
        SymPyDeprecationWarning(
            feature="'dummy=True' to create numbered Dummy symbols",
            useinstead="cls=Dummy",
            issue=3378, deprecated_since_version="0.7.0",
        ).warn()
        if assumptions.pop('dummy'):
            cls = C.Dummy

    if cls is None:
        # We can't just make the default cls=C.Symbol because it isn't
        # imported yet.
        cls = C.Symbol

    while True:
        name = '%s%s' % (prefix, start)
        yield cls(name, *args, **assumptions)
        start += 1

def capture(func):
    """Return the printed output of func().

    `func` should be a function without arguments that produces output with
    print statements.

    >>> from sympy.utilities.iterables import capture
    >>> from sympy import pprint
    >>> from sympy.abc import x
    >>> def foo():
    ...     print('hello world!')
    ...
    >>> 'hello' in capture(foo) # foo, not foo()
    True
    >>> capture(lambda: pprint(2/x))
    '2\\n-\\nx\\n'

    """
    import io
    import sys

    stdout = sys.stdout
    sys.stdout = file = io.StringIO()
    try:
        func()
    finally:
        sys.stdout = stdout
    return file.getvalue()

def sift(expr, keyfunc):
    """
    Sift the arguments of expr into a dictionary according to keyfunc.

    INPUT: expr may be an expression or iterable; if it is an expr then
    it is converted to a list of expr's args or [expr] if there are no args.

    OUTPUT: each element in expr is stored in a list keyed to the value
    of keyfunc for the element.

    Note that for a SymPy expression, the order of the elements in the lists
    is dependent on the order in .args, which can be arbitrary.

    Examples
    ========

    >>> from sympy.utilities import sift
    >>> from sympy.abc import x, y
    >>> from sympy import sqrt, exp

    >>> sift(list(range(5)), lambda x: x%2)
    {0: [0, 2, 4], 1: [1, 3]}

    sift() returns a defaultdict() object, so any key that has no matches will
    give [].

    >>> sift(x, lambda x: x.is_commutative)
    {True: [x]}
    >>> _[False]
    []

    Sometimes you won't know how many keys you will get:

    >>> sift(sqrt(x) + exp(x) + (y**x)**2,
    ... lambda x: x.as_base_exp()[0])
    {E: [exp(x)], x: [sqrt(x)], y: [y**(2*x)]}

    If you need to sort the sifted items it might be better to use
    the lazyDSU_sort which can economically apply multiple sort keys
    to a squence while sorting.

    See Also
    ========
    lazyDSU_sort
    """
    d = defaultdict(list)
    if hasattr(expr, 'args'):
        expr = expr.args or [expr]
    for e in expr:
        d[keyfunc(e)].append(e)
    return d

def take(iter, n):
    """Return ``n`` items from ``iter`` iterator. """
    return [ value for _, value in zip(range(n), iter) ]

def dict_merge(*dicts):
    """Merge dictionaries into a single dictionary. """
    merged = {}

    for dict in dicts:
        merged.update(dict)

    return merged

def common_prefix(*seqs):
    """Return the subsequence that is a common start of sequences in ``seqs``.

    >>> from sympy.utilities.iterables import common_prefix
    >>> common_prefix(list(range(3)))
    [0, 1, 2]
    >>> common_prefix(list(range(3)), list(range(4)))
    [0, 1, 2]
    >>> common_prefix([1, 2, 3], [1, 2, 5])
    [1, 2]
    >>> common_prefix([1, 2, 3], [1, 3, 5])
    [1]
    """
    if any(not s for s in seqs):
        return []
    elif len(seqs) == 1:
        return seqs[0]
    i = 0
    for i in range(min(len(s) for s in seqs)):
        if not all(seqs[j][i] == seqs[0][i] for j in range(len(seqs))):
            break
    else:
        i += 1
    return seqs[0][:i]

def common_suffix(*seqs):
    """Return the subsequence that is a common ending of sequences in ``seqs``.

    >>> from sympy.utilities.iterables import common_suffix
    >>> common_suffix(list(range(3)))
    [0, 1, 2]
    >>> common_suffix(list(range(3)), list(range(4)))
    []
    >>> common_suffix([1, 2, 3], [9, 2, 3])
    [2, 3]
    >>> common_suffix([1, 2, 3], [9, 7, 3])
    [3]
    """

    if any(not s for s in seqs):
        return []
    elif len(seqs) == 1:
        return seqs[0]
    i = 0
    for i in range(-1, -min(len(s) for s in seqs) - 1, -1):
        if not all(seqs[j][i] == seqs[0][i] for j in range(len(seqs))):
            break
    else:
        i -= 1
    if i == -1:
        return []
    else:
        return seqs[0][i + 1:]

def prefixes(seq):
    """
    Generate all prefixes of a sequence.

    Examples
    ========

    >>> from sympy.utilities.iterables import prefixes

    >>> list(prefixes([1,2,3,4]))
    [[1], [1, 2], [1, 2, 3], [1, 2, 3, 4]]

    """
    n = len(seq)

    for i in range(n):
        yield seq[:i+1]

def postfixes(seq):
    """
    Generate all postfixes of a sequence.

    Examples
    ========

    >>> from sympy.utilities.iterables import postfixes

    >>> list(postfixes([1,2,3,4]))
    [[4], [3, 4], [2, 3, 4], [1, 2, 3, 4]]

    """
    n = len(seq)

    for i in range(n):
        yield seq[n-i-1:]

def topological_sort(graph, key=None):
    r"""
    Topological sort of graph's vertices.

    **Parameters**

    ``graph`` : ``tuple[list, list[tuple[T, T]]``
        A tuple consisting of a list of vertices and a list of edges of
        a graph to be sorted topologically.

    ``key`` : ``callable[T]`` (optional)
        Ordering key for vertices on the same level. By default the natural
        (e.g. lexicographic) ordering is used (in this case the base type
        must implement ordering relations).

    Examples
    ========

    Consider a graph::

        +---+     +---+     +---+
        | 7 |\    | 5 |     | 3 |
        +---+ \   +---+     +---+
          |   _\___/ ____   _/ |
          |  /  \___/    \ /   |
          V  V           V V   |
         +----+         +---+  |
         | 11 |         | 8 |  |
         +----+         +---+  |
          | | \____   ___/ _   |
          | \      \ /    / \  |
          V  \     V V   /  V  V
        +---+ \   +---+ |  +----+
        | 2 |  |  | 9 | |  | 10 |
        +---+  |  +---+ |  +----+
               \________/

    where vertices are integers. This graph can be encoded using
    elementary Python's data structures as follows::

        >>> V = [2, 3, 5, 7, 8, 9, 10, 11]
        >>> E = [(7, 11), (7, 8), (5, 11), (3, 8), (3, 10),
        ...      (11, 2), (11, 9), (11, 10), (8, 9)]

    To compute a topological sort for graph ``(V, E)`` issue::

        >>> from sympy.utilities.iterables import topological_sort

        >>> topological_sort((V, E))
        [3, 5, 7, 8, 11, 2, 9, 10]

    If specific tie breaking approach is needed, use ``key`` parameter::

        >>> topological_sort((V, E), key=lambda v: -v)
        [7, 5, 11, 3, 10, 8, 9, 2]

    Only acyclic graphs can be sorted. If the input graph has a cycle,
    then :py:exc:`ValueError` will be raised::

        >>> topological_sort((V, E + [(10, 7)]))
        Traceback (most recent call last):
        ...
        ValueError: cycle detected

    .. seealso:: http://en.wikipedia.org/wiki/Topological_sorting

    """
    V, E = graph

    L = []
    S = set(V)
    E = list(E)

    for v, u in E:
        S.discard(u)

    if key is None:
        key = lambda value: value

    S = sorted(S, key=key, reverse=True)

    while S:
        node = S.pop()
        L.append(node)

        for u, v in list(E):
            if u == node:
                E.remove((u, v))

                for _u, _v in E:
                    if v == _v:
                        break
                else:
                    kv = key(v)

                    for i, s in enumerate(S):
                        ks = key(s)

                        if kv > ks:
                            S.insert(i, v)
                            break
                    else:
                        S.append(v)

    if E:
        raise ValueError("cycle detected")
    else:
        return L

def rotate_left(x, y):
    """
    Left rotates a list x by the number of steps specified
    in y.

    Examples
    ========

    >>> from sympy.utilities.iterables import rotate_left
    >>> a = [0, 1, 2]
    >>> rotate_left(a, 1)
    [1, 2, 0]
    """
    if len(x) == 0:
        return x
    y = y % len(x)
    return x[y:] + x[:y]

def rotate_right(x, y):
    """
    Left rotates a list x by the number of steps specified
    in y.

    Examples
    ========

    >>> from sympy.utilities.iterables import rotate_right
    >>> a = [0, 1, 2]
    >>> rotate_right(a, 1)
    [2, 0, 1]
    """
    if len(x) == 0:
        return x
    y = len(x) - y % len(x)
    return x[y:] + x[:y]

def multiset_partitions(multiset, m):
    """
    This is the algorithm for generating multiset partitions
    as described by Knuth in TAOCP Vol 4.

    Given a multiset, this algorithm visits all of its
    m-partitions, that is, all partitions having exactly size m
    using auxiliary arrays as described in the book.

    Examples
    ========

    >>> from sympy.utilities.iterables import multiset_partitions
    >>> list(multiset_partitions([1,2,3,4], 2))
    [[[1, 2, 3], [4]], [[1, 3], [2, 4]], [[1], [2, 3, 4]], [[1, 2], \
    [3, 4]], [[1, 2, 4], [3]], [[1, 4], [2, 3]], [[1, 3, 4], [2]]]
    >>> list(multiset_partitions([1,2,3,4], 1))
    [[[1, 2, 3, 4]]]
    >>> list(multiset_partitions([1,2,3,4], 4))
    [[[1], [2], [3], [4]]]

    See Also
    ========
    sympy.combinatorics.partitions.Partition
    sympy.combinatorics.partitions.IntegerPartition
    """
    cache = {}

    def visit(n, a):
        ps = [[] for i in range(m)]
        for j in range(n):
            ps[a[j + 1]].append(multiset[j])
        canonical = tuple(tuple(j) for j in ps)
        if not canonical in cache:
            cache[canonical] = 1
            return ps

    def f(m_arr, n_arr, sigma, n, a):
        if m_arr <= 2:
            v = visit(n, a)
            if v is not None:
                yield v
        else:
            for v in f(m_arr - 1, n_arr - 1, (m_arr + sigma) % 2, n, a):
                yield v
        if n_arr == m_arr + 1:
            a[m_arr] = m_arr - 1
            v = visit(n, a)
            if v is not None:
                yield v
            while a[n_arr] > 0:
                a[n_arr] = a[n_arr] - 1
                v = visit(n, a)
                if v is not None:
                    yield v
        elif n_arr > m_arr + 1:
            if (m_arr + sigma) % 2 == 1:
                a[n_arr - 1] = m_arr - 1
            else:
                a[m_arr] = m_arr - 1
            func = [f, b][(a[n_arr] + sigma) % 2]
            for v in func(m_arr, n_arr - 1, 0, n, a):
                if v is not None:
                    yield v
            while a[n_arr] > 0:
                a[n_arr] = a[n_arr] - 1
                func = [f, b][(a[n_arr] + sigma) % 2]
                for v in func(m_arr, n_arr - 1, 0, n, a):
                    if v is not None:
                        yield v

    def b(m_arr, n_arr, sigma, n, a):
        if n_arr == m_arr + 1:
            v = visit(n, a)
            if v is not None:
                yield v
            while a[n_arr] < m_arr - 1:
                a[n_arr] = a[n_arr] + 1
                v = visit(n, a)
                if v is not None:
                    yield v
            a[m_arr] = 0
            v = visit(n, a)
            if v is not None:
                yield v
        elif n_arr > m_arr + 1:
            func = [f, b][(a[n_arr] + sigma) % 2]
            for v in func(m_arr, n_arr - 1, 0, n, a):
                if v is not None:
                    yield v
            while a[n_arr] < m_arr - 1:
                a[n_arr] = a[n_arr] + 1
                func = [f, b][(a[n_arr] + sigma) % 2]
                for v in func(m_arr, n_arr - 1, 0, n, a):
                    if v is not None:
                        yield v
            if (m_arr + sigma) % 2 == 1:
                a[n_arr - 1] = 0
            else:
                a[m_arr] = 0
        if m_arr <= 2:
            v = visit(n, a)
            if v is not None:
                yield v
        else:
            for v in b(m_arr - 1, n_arr - 1, (m_arr + sigma) % 2, n, a):
                if v is not None:
                    yield v

    n = len(multiset)
    a = [0] * (n + 1)
    for j in range(1, m + 1):
        a[n - m + j] = j - 1
    return f(m, n, 0, n, a)

def partitions(n, m=None, k=None):
    """Generate all partitions of integer n (>= 0).

    'm' limits the number of parts in the partition, e.g. if m=2 then
        partitions will contain no more than 2 numbers, while
    'k' limits the numbers which may appear in the partition, e.g. k=2 will
        return partitions with no element greater than 2.

    Each partition is represented as a dictionary, mapping an integer
    to the number of copies of that integer in the partition.  For example,
    the first partition of 4 returned is {4: 1}: a single 4.

    >>> from sympy.utilities.iterables import partitions

    Maximum key (number in partition) limited with k (in this case, 2):

    >>> for p in partitions(6, k=2):
    ...     print(p)
    {2: 3}
    {1: 2, 2: 2}
    {1: 4, 2: 1}
    {1: 6}

    Maximum number of parts in partion limited with m (in this case, 2):

    >>> for p in partitions(6, m=2):
    ...     print(p)
    ...
    {6: 1}
    {1: 1, 5: 1}
    {2: 1, 4: 1}
    {3: 2}

    Note that the _same_ dictionary object is returned each time.
    This is for speed:  generating each partition goes quickly,
    taking constant time independent of n.

    >>> [p for p in partitions(6, k=2)]
    [{1: 6}, {1: 6}, {1: 6}, {1: 6}]

    If you want to build a list of the returned dictionaries then
    make a copy of them:

    >>> [p.copy() for p in partitions(6, k=2)]
    [{2: 3}, {1: 2, 2: 2}, {1: 4, 2: 1}, {1: 6}]

    Reference:
        modified from Tim Peter's version to allow for k and m values:
        code.activestate.com/recipes/218332-generator-for-integer-partitions/

    See Also
    ========
    sympy.combinatorics.partitions.Partition
    sympy.combinatorics.partitions.IntegerPartition
    """
    if n < 0:
        raise ValueError("n must be >= 0")
    if m == 0:
        raise ValueError("m must be > 0")
    m = min(m or n, n)
    if m < 1:
        raise ValueError("maximum numbers in partition, m, must be > 0")
    k = min(k or n, n)
    if k < 1:
        raise ValueError("maximum value in partition, k, must be > 0")

    if m*k < n:
        return

    n, m, k = as_int(n), as_int(m), as_int(k)
    q, r = divmod(n, k)
    ms = {k: q}
    keys = [k]  # ms.keys(), from largest to smallest
    if r:
        ms[r] = 1
        keys.append(r)
    room = m - q - bool(r)
    yield ms

    while keys != [1]:
        # Reuse any 1's.
        if keys[-1] == 1:
            del keys[-1]
            reuse = ms.pop(1)
            room += reuse
        else:
            reuse = 0

        while 1:
            # Let i be the smallest key larger than 1.  Reuse one
            # instance of i.
            i = keys[-1]
            newcount = ms[i] = ms[i] - 1
            reuse += i
            if newcount == 0:
                del keys[-1], ms[i]
            room += 1


            # Break the remainder into pieces of size i-1.
            i -= 1
            q, r = divmod(reuse, i)
            need = q + bool(r)
            if need > room:
                if not keys:
                    return
                continue

            ms[i] = q
            keys.append(i)
            if r:
                ms[r] = 1
                keys.append(r)
            break
        room -= need
        yield ms

def binary_partitions(n):
    """
    Generates the binary partition of n.

    A binary partition consists only of numbers that are
    powers of two. Each step reduces a 2**(k+1) to 2**k and
    2**k. Thus 16 is converted to 8 and 8.

    Reference: TAOCP 4, section 7.2.1.5, problem 64

    Examples
    ========

    >>> from sympy.utilities.iterables import binary_partitions
    >>> for i in binary_partitions(5):
    ...     print(i)
    ...
    [4, 1]
    [2, 2, 1]
    [2, 1, 1, 1]
    [1, 1, 1, 1, 1]
    """
    from math import ceil, log
    pow = int(2**(ceil(log(n, 2))))
    sum = 0
    partition = []
    while pow:
        if sum + pow <= n:
            partition.append(pow)
            sum += pow
        pow >>= 1

    last_num = len(partition) - 1 - (n & 1)
    while last_num >= 0:
        yield partition
        if partition[last_num] == 2:
            partition[last_num] = 1
            partition.append(1)
            last_num -= 1
            continue
        partition.append(1)
        partition[last_num] >>= 1
        x = partition[last_num + 1] = partition[last_num]
        last_num += 1
        while x > 1:
            if x <= len(partition) - last_num - 1:
                del partition[-x + 1:]
                last_num += 1
                partition[last_num] = x
            else:
                x >>= 1
    yield [1]*n

def has_dups(seq):
    """Return True if there are any duplicate elements in ``seq``.

    Examples
    ========
    >>> from sympy.utilities.iterables import has_dups
    >>> from sympy import Dict, Set

    >>> has_dups((1, 2, 1))
    True
    >>> has_dups(list(range(3)))
    False
    >>> all(has_dups(c) is False for c in (set(), Set(), dict(), Dict()))
    True
    """
    if isinstance(seq, (dict, set, C.Dict, C.Set)):
        return False
    uniq = set()
    return any(True for s in seq if s in uniq or uniq.add(s))

def has_variety(seq):
    """Return True if there are any different elements in ``seq``.

    Examples
    ========
    >>> from sympy.utilities.iterables import has_variety
    >>> from sympy import Dict, Set

    >>> has_variety((1, 2, 1))
    True
    >>> has_variety((1, 1, 1))
    False
    """
    for i, s in enumerate(seq):
        if i == 0:
            sentinel = s
        else:
            if s != sentinel:
                return True
    return False

def uniq(seq):
    """
    Remove repeated elements from an iterable, preserving order of first
    appearance.

    Returns a sequence of the same type of the input, or a list if the input
    was not a sequence.

    Examples
    ========

    >>> from sympy.utilities.iterables import uniq
    >>> uniq([1,4,1,5,4,2,1,2])
    [1, 4, 5, 2]
    >>> uniq((1,4,1,5,4,2,1,2))
    (1, 4, 5, 2)
    >>> uniq(x for x in (1,4,1,5,4,2,1,2))
    [1, 4, 5, 2]

    """
    from sympy.core.function import Tuple
    seen = set()
    result = (s for s in seq if not (s in seen or seen.add(s)))
    if not hasattr(seq, '__getitem__'):
        return list(result)
    if isinstance(seq, Tuple):
        return Tuple(*tuple(result))
    return type(seq)(result)

def generate_bell(n):
    """
    Generates the bell permutations.

    In a Bell permutation, each cycle is a decreasing
    sequence of integers.

    Reference:
    [1] Generating involutions, derangements, and relatives by ECO
    Vincent Vajnovszki, DMTCS vol 1 issue 12, 2010

    Examples
    ========

    >>> from sympy.utilities.iterables import generate_bell
    >>> list(generate_bell(3))
    [(0, 1, 2), (0, 2, 1), (1, 0, 2), (2, 0, 1), (2, 1, 0)]
    """
    P = [i for i in range(n)]
    T = [0]
    cache = set()
    def gen(P, T, t):
        if t == (n - 1):
            cache.add(tuple(P))
        else:
            for i in T:
                P[i], P[t+1] = P[t+1], P[i]
                if tuple(P) not in cache:
                    cache.add(tuple(P))
                    gen(P, T, t + 1)
                P[i], P[t+1] = P[t+1], P[i]
            T.append(t + 1)
            cache.add(tuple(P))
            gen(P, T, t + 1)
            T.remove(t + 1)
    gen(P, T, 0)
    return sorted(cache)

def generate_involutions(n):
    """
    Generates involutions.

    An involution is a permutation that when multiplied
    by itself equals the identity permutation. In this
    implementation the involutions are generated using
    Fixed Points.

    Alternatively, an involution can be considered as
    a permutation that does not contain any cycles with
    a length that is greater than two.

    Reference:
    http://mathworld.wolfram.com/PermutationInvolution.html

    Examples
    ========

    >>> from sympy.utilities.iterables import generate_involutions
    >>> generate_involutions(3)
    [(0, 1, 2), (0, 2, 1), (1, 0, 2), (2, 1, 0)]
    >>> len(generate_involutions(4))
    10
    """
    P = list(range(n)) # the items of the permutation
    F = [0] # the fixed points {is this right??}
    cache = set()
    def gen(P, F, t):
        if t == n:
            cache.add(tuple(P))
        else:
            for j in range(len(F)):
                P[j], P[t] = P[t], P[j]
                if tuple(P) not in cache:
                    cache.add(tuple(P))
                    Fj = F.pop(j)
                    gen(P, F, t + 1)
                    F.insert(j, Fj)
                P[j], P[t] = P[t], P[j]
            t += 1
            F.append(t)
            cache.add(tuple(P))
            gen(P, F, t)
            F.pop()
    gen(P, F, 1)
    return sorted(cache)

def generate_derangements(perm):
    """
    Routine to generate derangements.

    TODO: This will be rewritten to use the
    ECO operator approach once the permutations
    branch is in master.

    Examples
    ========

    >>> from sympy.utilities.iterables import generate_derangements
    >>> list(generate_derangements([0,1,2]))
    [[1, 2, 0], [2, 0, 1]]
    >>> list(generate_derangements([0,1,2,3]))
    [[1, 0, 3, 2], [1, 2, 3, 0], [1, 3, 0, 2], [2, 0, 3, 1], \
    [2, 3, 0, 1], [2, 3, 1, 0], [3, 0, 1, 2], [3, 2, 0, 1], \
    [3, 2, 1, 0]]
    >>> list(generate_derangements([0,1,1]))
    []
    """
    indices = list(range(len(perm)))
    p = variations(indices, len(indices))
    for rv in \
            uniq(tuple(perm[i] for i in idx) \
                 for idx in p if all(perm[k] != \
                                     perm[idx[k]] for k in range(len(perm)))):
        yield list(rv)

def unrestricted_necklace(n, k):
    """
    A routine to generate unrestriced necklaces.

    Here n is the length of the necklace and k - 1
    is the maximum permissible element in the
    generated necklaces.

    Reference:
    http://mathworld.wolfram.com/Necklace.html

    Examples
    ========

    >>> from sympy.utilities.iterables import unrestricted_necklace
    >>> [i[:] for i in unrestricted_necklace(3, 2)]
    [[0, 0, 0], [0, 1, 1]]
    >>> [i[:] for i in unrestricted_necklace(4, 4)]
    [[0, 0, 0, 0], [0, 0, 1, 0], [0, 0, 2, 0], [0, 0, 3, 0], \
    [0, 1, 1, 1], [0, 1, 2, 1], [0, 1, 3, 1], [0, 2, 2, 2], \
    [0, 2, 3, 2], [0, 3, 3, 3]]
    """
    a = [0] * n
    def gen(t, p):
        if (t > n - 1):
            if (n % p == 0):
                yield a
        else:
            a[t] = a[t - p]
            for necklace in gen(t + 1, p):
                yield necklace
            for j in range(a[t - p] + 1, k):
                a[t] = j
                for necklace in gen(t + 1, t):
                    yield necklace
    return gen(1, 1)

def generate_oriented_forest(n):
    """
    This algorithm generates oriented forests.

    An oriented graph is a directed graph having no symmetric pair of directed
    edges. A forest is an acyclic graph, i.e., it has no cycles. A forest can
    also be described as a disjoint union of trees, which are graphs in which
    any two vertices are connected by exactly one simple path.

    Reference:
    [1] T. Beyer and S.M. Hedetniemi: constant time generation of \
        rooted trees, SIAM J. Computing Vol. 9, No. 4, November 1980
    [2] http://stackoverflow.com/questions/1633833/oriented-forest-taocp-algorithm-in-python

    Examples
    ========

    >>> from sympy.utilities.iterables import generate_oriented_forest
    >>> list(generate_oriented_forest(4))
    [[0, 1, 2, 3], [0, 1, 2, 2], [0, 1, 2, 1], [0, 1, 2, 0], \
    [0, 1, 1, 1], [0, 1, 1, 0], [0, 1, 0, 1], [0, 1, 0, 0], [0, 0, 0, 0]]
    """
    P = list(range(-1, n))
    while True:
        yield P[1:]
        if P[n] > 0:
            P[n] = P[P[n]]
        else:
            for p in range(n - 1, 0, -1):
                if P[p] != 0:
                    target = P[p] - 1
                    for q in range(p - 1, 0, -1):
                        if P[q] == target:
                            break
                    offset = p - q
                    for i in range(p, n + 1):
                        P[i] = P[i - offset]
                    break
            else:
                break

def lazyDSU_sort(seq, keys, warn=False):
    """Return sorted seq, breaking ties by applying keys only when needed.

    If ``warn`` is True then an error will be raised if there were no
    keys remaining to break ties.

    Examples
    ========

    >>> from sympy.utilities.iterables import lazyDSU_sort, default_sort_key
    >>> from sympy.abc import x, y

    The count_ops is not sufficient to break ties in this list and the first
    two items appear in their original order: Python sorting is stable.

    >>> lazyDSU_sort([y + 2, x + 2, x**2 + y + 3],
    ...    [lambda x: x.count_ops()], warn=False)
    ...
    [y + 2, x + 2, x**2 + y + 3]

    The use of default_sort_key allows the tie to be broken (and warn can
    be safely left False).

    >>> lazyDSU_sort([y + 2, x + 2, x**2 + y + 3],
    ...    [lambda x: x.count_ops(),
    ...     default_sort_key])
    ...
    [x + 2, y + 2, x**2 + y + 3]

    Here, sequences are sorted by length, then sum:

    >>> seq, keys = [[[1, 2, 1], [0, 3, 1], [1, 1, 3], [2], [1]], [
    ...    lambda x: len(x),
    ...    lambda x: sum(x)]]
    ...
    >>> lazyDSU_sort(seq, keys, warn=False)
    [[1], [2], [1, 2, 1], [0, 3, 1], [1, 1, 3]]

    If ``warn`` is True, an error will be raised if there were not
    enough keys to break ties:

    >>> lazyDSU_sort(seq, keys, warn=True)
    Traceback (most recent call last):
    ...
    ValueError: not enough keys to break ties


    Notes
    =====

    The decorated sort is one of the fastest ways to sort a sequence for
    which special item comparison is desired: the sequence is decorated,
    sorted on the basis of the decoration (e.g. making all letters lower
    case) and then undecorated. If one wants to break ties for items that
    have the same decorated value, a second key can be used. But if the
    second key is expensive to compute then it is inefficient to decorate
    all items with both keys: only those items having identical first key
    values need to be decorated. This function applies keys successively
    only when needed to break ties.
    """
    d = defaultdict(list)
    keys = list(keys)
    f = keys.pop(0)
    for a in seq:
        d[f(a)].append(a)
    seq = []
    for k in sorted(d.keys()):
        if len(d[k]) > 1:
            if keys:
                d[k] = lazyDSU_sort(d[k], keys, warn)
            elif warn:
              raise ValueError('not enough keys to break ties')
            seq.extend(d[k])
        else:
            seq.append(d[k][0])
    return seq

def minlex(seq, directed=True):
    """Return a tuple where the smallest element appears first; if
    ``directed`` is True (default) then the order is preserved, otherwise the
    sequence will be reversed if that gives a lexically smaller ordering.

    Examples
    ========
    >>> from sympy.combinatorics.polyhedron import minlex
    >>> minlex((1, 2, 0))
    (0, 1, 2)
    >>> minlex((1, 0, 2))
    (0, 2, 1)
    >>> minlex((1, 0, 2), directed=False)
    (0, 1, 2)
    """
    seq = list(seq)
    small = min(seq)
    i = seq.index(small)
    if not directed:
        n = len(seq)
        p = (i + 1) % n
        m = (i - 1) % n
        if seq[p] > seq[m]:
            seq = list(reversed(seq))
            i = n - i - 1
    if i:
        seq = rotate_left(seq, i)
    return tuple(seq)

def runs(seq, op=gt):
    """Group the sequence into lists in which successive elements
    all compare the same with the comparison operator, ``op``:
    op(seq[i + 1], seq[i]) is True from all elements in a run.

    Examples
    ========

    >>> from sympy.utilities.iterables import runs
    >>> from operator import ge
    >>> runs([0, 1, 2, 2, 1, 4, 3, 2, 2])
    [[0, 1, 2], [2], [1, 4], [3], [2], [2]]
    >>> runs([0, 1, 2, 2, 1, 4, 3, 2, 2], op=ge)
    [[0, 1, 2, 2], [1, 4], [3], [2, 2]]
    """
    cycles = []
    seq = iter(seq)
    try:
        run = [next(seq)]
    except StopIteration:
        return []
    while True:
        try:
            ei = next(seq)
        except StopIteration:
            break
        if op(ei, run[-1]):
            run.append(ei)
            continue
        else:
            cycles.append(run)
            run = [ei]
    if run:
        cycles.append(run)
    return cycles
