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diff --git a/contrib/tools/python3/Lib/heapq.py b/contrib/tools/python3/Lib/heapq.py
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+++ b/contrib/tools/python3/Lib/heapq.py
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+"""Heap queue algorithm (a.k.a. priority queue).
+
+Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
+all k, counting elements from 0.  For the sake of comparison,
+non-existing elements are considered to be infinite.  The interesting
+property of a heap is that a[0] is always its smallest element.
+
+Usage:
+
+heap = []            # creates an empty heap
+heappush(heap, item) # pushes a new item on the heap
+item = heappop(heap) # pops the smallest item from the heap
+item = heap[0]       # smallest item on the heap without popping it
+heapify(x)           # transforms list into a heap, in-place, in linear time
+item = heappushpop(heap, item) # pushes a new item and then returns
+                               # the smallest item; the heap size is unchanged
+item = heapreplace(heap, item) # pops and returns smallest item, and adds
+                               # new item; the heap size is unchanged
+
+Our API differs from textbook heap algorithms as follows:
+
+- We use 0-based indexing.  This makes the relationship between the
+  index for a node and the indexes for its children slightly less
+  obvious, but is more suitable since Python uses 0-based indexing.
+
+- Our heappop() method returns the smallest item, not the largest.
+
+These two make it possible to view the heap as a regular Python list
+without surprises: heap[0] is the smallest item, and heap.sort()
+maintains the heap invariant!
+"""
+
+# Original code by Kevin O'Connor, augmented by Tim Peters and Raymond Hettinger
+
+__about__ = """Heap queues
+
+[explanation by François Pinard]
+
+Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
+all k, counting elements from 0.  For the sake of comparison,
+non-existing elements are considered to be infinite.  The interesting
+property of a heap is that a[0] is always its smallest element.
+
+The strange invariant above is meant to be an efficient memory
+representation for a tournament.  The numbers below are `k', not a[k]:
+
+                                   0
+
+                  1                                 2
+
+          3               4                5               6
+
+      7       8       9       10      11      12      13      14
+
+    15 16   17 18   19 20   21 22   23 24   25 26   27 28   29 30
+
+
+In the tree above, each cell `k' is topping `2*k+1' and `2*k+2'.  In
+a usual binary tournament we see in sports, each cell is the winner
+over the two cells it tops, and we can trace the winner down the tree
+to see all opponents s/he had.  However, in many computer applications
+of such tournaments, we do not need to trace the history of a winner.
+To be more memory efficient, when a winner is promoted, we try to
+replace it by something else at a lower level, and the rule becomes
+that a cell and the two cells it tops contain three different items,
+but the top cell "wins" over the two topped cells.
+
+If this heap invariant is protected at all time, index 0 is clearly
+the overall winner.  The simplest algorithmic way to remove it and
+find the "next" winner is to move some loser (let's say cell 30 in the
+diagram above) into the 0 position, and then percolate this new 0 down
+the tree, exchanging values, until the invariant is re-established.
+This is clearly logarithmic on the total number of items in the tree.
+By iterating over all items, you get an O(n ln n) sort.
+
+A nice feature of this sort is that you can efficiently insert new
+items while the sort is going on, provided that the inserted items are
+not "better" than the last 0'th element you extracted.  This is
+especially useful in simulation contexts, where the tree holds all
+incoming events, and the "win" condition means the smallest scheduled
+time.  When an event schedule other events for execution, they are
+scheduled into the future, so they can easily go into the heap.  So, a
+heap is a good structure for implementing schedulers (this is what I
+used for my MIDI sequencer :-).
+
+Various structures for implementing schedulers have been extensively
+studied, and heaps are good for this, as they are reasonably speedy,
+the speed is almost constant, and the worst case is not much different
+than the average case.  However, there are other representations which
+are more efficient overall, yet the worst cases might be terrible.
+
+Heaps are also very useful in big disk sorts.  You most probably all
+know that a big sort implies producing "runs" (which are pre-sorted
+sequences, which size is usually related to the amount of CPU memory),
+followed by a merging passes for these runs, which merging is often
+very cleverly organised[1].  It is very important that the initial
+sort produces the longest runs possible.  Tournaments are a good way
+to that.  If, using all the memory available to hold a tournament, you
+replace and percolate items that happen to fit the current run, you'll
+produce runs which are twice the size of the memory for random input,
+and much better for input fuzzily ordered.
+
+Moreover, if you output the 0'th item on disk and get an input which
+may not fit in the current tournament (because the value "wins" over
+the last output value), it cannot fit in the heap, so the size of the
+heap decreases.  The freed memory could be cleverly reused immediately
+for progressively building a second heap, which grows at exactly the
+same rate the first heap is melting.  When the first heap completely
+vanishes, you switch heaps and start a new run.  Clever and quite
+effective!
+
+In a word, heaps are useful memory structures to know.  I use them in
+a few applications, and I think it is good to keep a `heap' module
+around. :-)
+
+--------------------
+[1] The disk balancing algorithms which are current, nowadays, are
+more annoying than clever, and this is a consequence of the seeking
+capabilities of the disks.  On devices which cannot seek, like big
+tape drives, the story was quite different, and one had to be very
+clever to ensure (far in advance) that each tape movement will be the
+most effective possible (that is, will best participate at
+"progressing" the merge).  Some tapes were even able to read
+backwards, and this was also used to avoid the rewinding time.
+Believe me, real good tape sorts were quite spectacular to watch!
+From all times, sorting has always been a Great Art! :-)
+"""
+
+__all__ = ['heappush', 'heappop', 'heapify', 'heapreplace', 'merge',
+           'nlargest', 'nsmallest', 'heappushpop']
+
+def heappush(heap, item):
+    """Push item onto heap, maintaining the heap invariant."""
+    heap.append(item)
+    _siftdown(heap, 0, len(heap)-1)
+
+def heappop(heap):
+    """Pop the smallest item off the heap, maintaining the heap invariant."""
+    lastelt = heap.pop()    # raises appropriate IndexError if heap is empty
+    if heap:
+        returnitem = heap[0]
+        heap[0] = lastelt
+        _siftup(heap, 0)
+        return returnitem
+    return lastelt
+
+def heapreplace(heap, item):
+    """Pop and return the current smallest value, and add the new item.
+
+    This is more efficient than heappop() followed by heappush(), and can be
+    more appropriate when using a fixed-size heap.  Note that the value
+    returned may be larger than item!  That constrains reasonable uses of
+    this routine unless written as part of a conditional replacement:
+
+        if item > heap[0]:
+            item = heapreplace(heap, item)
+    """
+    returnitem = heap[0]    # raises appropriate IndexError if heap is empty
+    heap[0] = item
+    _siftup(heap, 0)
+    return returnitem
+
+def heappushpop(heap, item):
+    """Fast version of a heappush followed by a heappop."""
+    if heap and heap[0] < item:
+        item, heap[0] = heap[0], item
+        _siftup(heap, 0)
+    return item
+
+def heapify(x):
+    """Transform list into a heap, in-place, in O(len(x)) time."""
+    n = len(x)
+    # Transform bottom-up.  The largest index there's any point to looking at
+    # is the largest with a child index in-range, so must have 2*i + 1 < n,
+    # or i < (n-1)/2.  If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so
+    # j-1 is the largest, which is n//2 - 1.  If n is odd = 2*j+1, this is
+    # (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1.
+    for i in reversed(range(n//2)):
+        _siftup(x, i)
+
+def _heappop_max(heap):
+    """Maxheap version of a heappop."""
+    lastelt = heap.pop()    # raises appropriate IndexError if heap is empty
+    if heap:
+        returnitem = heap[0]
+        heap[0] = lastelt
+        _siftup_max(heap, 0)
+        return returnitem
+    return lastelt
+
+def _heapreplace_max(heap, item):
+    """Maxheap version of a heappop followed by a heappush."""
+    returnitem = heap[0]    # raises appropriate IndexError if heap is empty
+    heap[0] = item
+    _siftup_max(heap, 0)
+    return returnitem
+
+def _heapify_max(x):
+    """Transform list into a maxheap, in-place, in O(len(x)) time."""
+    n = len(x)
+    for i in reversed(range(n//2)):
+        _siftup_max(x, i)
+
+# 'heap' is a heap at all indices >= startpos, except possibly for pos.  pos
+# is the index of a leaf with a possibly out-of-order value.  Restore the
+# heap invariant.
+def _siftdown(heap, startpos, pos):
+    newitem = heap[pos]
+    # Follow the path to the root, moving parents down until finding a place
+    # newitem fits.
+    while pos > startpos:
+        parentpos = (pos - 1) >> 1
+        parent = heap[parentpos]
+        if newitem < parent:
+            heap[pos] = parent
+            pos = parentpos
+            continue
+        break
+    heap[pos] = newitem
+
+# The child indices of heap index pos are already heaps, and we want to make
+# a heap at index pos too.  We do this by bubbling the smaller child of
+# pos up (and so on with that child's children, etc) until hitting a leaf,
+# then using _siftdown to move the oddball originally at index pos into place.
+#
+# We *could* break out of the loop as soon as we find a pos where newitem <=
+# both its children, but turns out that's not a good idea, and despite that
+# many books write the algorithm that way.  During a heap pop, the last array
+# element is sifted in, and that tends to be large, so that comparing it
+# against values starting from the root usually doesn't pay (= usually doesn't
+# get us out of the loop early).  See Knuth, Volume 3, where this is
+# explained and quantified in an exercise.
+#
+# Cutting the # of comparisons is important, since these routines have no
+# way to extract "the priority" from an array element, so that intelligence
+# is likely to be hiding in custom comparison methods, or in array elements
+# storing (priority, record) tuples.  Comparisons are thus potentially
+# expensive.
+#
+# On random arrays of length 1000, making this change cut the number of
+# comparisons made by heapify() a little, and those made by exhaustive
+# heappop() a lot, in accord with theory.  Here are typical results from 3
+# runs (3 just to demonstrate how small the variance is):
+#
+# Compares needed by heapify     Compares needed by 1000 heappops
+# --------------------------     --------------------------------
+# 1837 cut to 1663               14996 cut to 8680
+# 1855 cut to 1659               14966 cut to 8678
+# 1847 cut to 1660               15024 cut to 8703
+#
+# Building the heap by using heappush() 1000 times instead required
+# 2198, 2148, and 2219 compares:  heapify() is more efficient, when
+# you can use it.
+#
+# The total compares needed by list.sort() on the same lists were 8627,
+# 8627, and 8632 (this should be compared to the sum of heapify() and
+# heappop() compares):  list.sort() is (unsurprisingly!) more efficient
+# for sorting.
+
+def _siftup(heap, pos):
+    endpos = len(heap)
+    startpos = pos
+    newitem = heap[pos]
+    # Bubble up the smaller child until hitting a leaf.
+    childpos = 2*pos + 1    # leftmost child position
+    while childpos < endpos:
+        # Set childpos to index of smaller child.
+        rightpos = childpos + 1
+        if rightpos < endpos and not heap[childpos] < heap[rightpos]:
+            childpos = rightpos
+        # Move the smaller child up.
+        heap[pos] = heap[childpos]
+        pos = childpos
+        childpos = 2*pos + 1
+    # The leaf at pos is empty now.  Put newitem there, and bubble it up
+    # to its final resting place (by sifting its parents down).
+    heap[pos] = newitem
+    _siftdown(heap, startpos, pos)
+
+def _siftdown_max(heap, startpos, pos):
+    'Maxheap variant of _siftdown'
+    newitem = heap[pos]
+    # Follow the path to the root, moving parents down until finding a place
+    # newitem fits.
+    while pos > startpos:
+        parentpos = (pos - 1) >> 1
+        parent = heap[parentpos]
+        if parent < newitem:
+            heap[pos] = parent
+            pos = parentpos
+            continue
+        break
+    heap[pos] = newitem
+
+def _siftup_max(heap, pos):
+    'Maxheap variant of _siftup'
+    endpos = len(heap)
+    startpos = pos
+    newitem = heap[pos]
+    # Bubble up the larger child until hitting a leaf.
+    childpos = 2*pos + 1    # leftmost child position
+    while childpos < endpos:
+        # Set childpos to index of larger child.
+        rightpos = childpos + 1
+        if rightpos < endpos and not heap[rightpos] < heap[childpos]:
+            childpos = rightpos
+        # Move the larger child up.
+        heap[pos] = heap[childpos]
+        pos = childpos
+        childpos = 2*pos + 1
+    # The leaf at pos is empty now.  Put newitem there, and bubble it up
+    # to its final resting place (by sifting its parents down).
+    heap[pos] = newitem
+    _siftdown_max(heap, startpos, pos)
+
+def merge(*iterables, key=None, reverse=False):
+    '''Merge multiple sorted inputs into a single sorted output.
+
+    Similar to sorted(itertools.chain(*iterables)) but returns a generator,
+    does not pull the data into memory all at once, and assumes that each of
+    the input streams is already sorted (smallest to largest).
+
+    >>> list(merge([1,3,5,7], [0,2,4,8], [5,10,15,20], [], [25]))
+    [0, 1, 2, 3, 4, 5, 5, 7, 8, 10, 15, 20, 25]
+
+    If *key* is not None, applies a key function to each element to determine
+    its sort order.
+
+    >>> list(merge(['dog', 'horse'], ['cat', 'fish', 'kangaroo'], key=len))
+    ['dog', 'cat', 'fish', 'horse', 'kangaroo']
+
+    '''
+
+    h = []
+    h_append = h.append
+
+    if reverse:
+        _heapify = _heapify_max
+        _heappop = _heappop_max
+        _heapreplace = _heapreplace_max
+        direction = -1
+    else:
+        _heapify = heapify
+        _heappop = heappop
+        _heapreplace = heapreplace
+        direction = 1
+
+    if key is None:
+        for order, it in enumerate(map(iter, iterables)):
+            try:
+                next = it.__next__
+                h_append([next(), order * direction, next])
+            except StopIteration:
+                pass
+        _heapify(h)
+        while len(h) > 1:
+            try:
+                while True:
+                    value, order, next = s = h[0]
+                    yield value
+                    s[0] = next()           # raises StopIteration when exhausted
+                    _heapreplace(h, s)      # restore heap condition
+            except StopIteration:
+                _heappop(h)                 # remove empty iterator
+        if h:
+            # fast case when only a single iterator remains
+            value, order, next = h[0]
+            yield value
+            yield from next.__self__
+        return
+
+    for order, it in enumerate(map(iter, iterables)):
+        try:
+            next = it.__next__
+            value = next()
+            h_append([key(value), order * direction, value, next])
+        except StopIteration:
+            pass
+    _heapify(h)
+    while len(h) > 1:
+        try:
+            while True:
+                key_value, order, value, next = s = h[0]
+                yield value
+                value = next()
+                s[0] = key(value)
+                s[2] = value
+                _heapreplace(h, s)
+        except StopIteration:
+            _heappop(h)
+    if h:
+        key_value, order, value, next = h[0]
+        yield value
+        yield from next.__self__
+
+
+# Algorithm notes for nlargest() and nsmallest()
+# ==============================================
+#
+# Make a single pass over the data while keeping the k most extreme values
+# in a heap.  Memory consumption is limited to keeping k values in a list.
+#
+# Measured performance for random inputs:
+#
+#                                   number of comparisons
+#    n inputs     k-extreme values  (average of 5 trials)   % more than min()
+# -------------   ----------------  ---------------------   -----------------
+#      1,000           100                  3,317               231.7%
+#     10,000           100                 14,046                40.5%
+#    100,000           100                105,749                 5.7%
+#  1,000,000           100              1,007,751                 0.8%
+# 10,000,000           100             10,009,401                 0.1%
+#
+# Theoretical number of comparisons for k smallest of n random inputs:
+#
+# Step   Comparisons                  Action
+# ----   --------------------------   ---------------------------
+#  1     1.66 * k                     heapify the first k-inputs
+#  2     n - k                        compare remaining elements to top of heap
+#  3     k * (1 + lg2(k)) * ln(n/k)   replace the topmost value on the heap
+#  4     k * lg2(k) - (k/2)           final sort of the k most extreme values
+#
+# Combining and simplifying for a rough estimate gives:
+#
+#        comparisons = n + k * (log(k, 2) * log(n/k) + log(k, 2) + log(n/k))
+#
+# Computing the number of comparisons for step 3:
+# -----------------------------------------------
+# * For the i-th new value from the iterable, the probability of being in the
+#   k most extreme values is k/i.  For example, the probability of the 101st
+#   value seen being in the 100 most extreme values is 100/101.
+# * If the value is a new extreme value, the cost of inserting it into the
+#   heap is 1 + log(k, 2).
+# * The probability times the cost gives:
+#            (k/i) * (1 + log(k, 2))
+# * Summing across the remaining n-k elements gives:
+#            sum((k/i) * (1 + log(k, 2)) for i in range(k+1, n+1))
+# * This reduces to:
+#            (H(n) - H(k)) * k * (1 + log(k, 2))
+# * Where H(n) is the n-th harmonic number estimated by:
+#            gamma = 0.5772156649
+#            H(n) = log(n, e) + gamma + 1 / (2 * n)
+#   http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Rate_of_divergence
+# * Substituting the H(n) formula:
+#            comparisons = k * (1 + log(k, 2)) * (log(n/k, e) + (1/n - 1/k) / 2)
+#
+# Worst-case for step 3:
+# ----------------------
+# In the worst case, the input data is reversed sorted so that every new element
+# must be inserted in the heap:
+#
+#             comparisons = 1.66 * k + log(k, 2) * (n - k)
+#
+# Alternative Algorithms
+# ----------------------
+# Other algorithms were not used because they:
+# 1) Took much more auxiliary memory,
+# 2) Made multiple passes over the data.
+# 3) Made more comparisons in common cases (small k, large n, semi-random input).
+# See the more detailed comparison of approach at:
+# http://code.activestate.com/recipes/577573-compare-algorithms-for-heapqsmallest
+
+def nsmallest(n, iterable, key=None):
+    """Find the n smallest elements in a dataset.
+
+    Equivalent to:  sorted(iterable, key=key)[:n]
+    """
+
+    # Short-cut for n==1 is to use min()
+    if n == 1:
+        it = iter(iterable)
+        sentinel = object()
+        result = min(it, default=sentinel, key=key)
+        return [] if result is sentinel else [result]
+
+    # When n>=size, it's faster to use sorted()
+    try:
+        size = len(iterable)
+    except (TypeError, AttributeError):
+        pass
+    else:
+        if n >= size:
+            return sorted(iterable, key=key)[:n]
+
+    # When key is none, use simpler decoration
+    if key is None:
+        it = iter(iterable)
+        # put the range(n) first so that zip() doesn't
+        # consume one too many elements from the iterator
+        result = [(elem, i) for i, elem in zip(range(n), it)]
+        if not result:
+            return result
+        _heapify_max(result)
+        top = result[0][0]
+        order = n
+        _heapreplace = _heapreplace_max
+        for elem in it:
+            if elem < top:
+                _heapreplace(result, (elem, order))
+                top, _order = result[0]
+                order += 1
+        result.sort()
+        return [elem for (elem, order) in result]
+
+    # General case, slowest method
+    it = iter(iterable)
+    result = [(key(elem), i, elem) for i, elem in zip(range(n), it)]
+    if not result:
+        return result
+    _heapify_max(result)
+    top = result[0][0]
+    order = n
+    _heapreplace = _heapreplace_max
+    for elem in it:
+        k = key(elem)
+        if k < top:
+            _heapreplace(result, (k, order, elem))
+            top, _order, _elem = result[0]
+            order += 1
+    result.sort()
+    return [elem for (k, order, elem) in result]
+
+def nlargest(n, iterable, key=None):
+    """Find the n largest elements in a dataset.
+
+    Equivalent to:  sorted(iterable, key=key, reverse=True)[:n]
+    """
+
+    # Short-cut for n==1 is to use max()
+    if n == 1:
+        it = iter(iterable)
+        sentinel = object()
+        result = max(it, default=sentinel, key=key)
+        return [] if result is sentinel else [result]
+
+    # When n>=size, it's faster to use sorted()
+    try:
+        size = len(iterable)
+    except (TypeError, AttributeError):
+        pass
+    else:
+        if n >= size:
+            return sorted(iterable, key=key, reverse=True)[:n]
+
+    # When key is none, use simpler decoration
+    if key is None:
+        it = iter(iterable)
+        result = [(elem, i) for i, elem in zip(range(0, -n, -1), it)]
+        if not result:
+            return result
+        heapify(result)
+        top = result[0][0]
+        order = -n
+        _heapreplace = heapreplace
+        for elem in it:
+            if top < elem:
+                _heapreplace(result, (elem, order))
+                top, _order = result[0]
+                order -= 1
+        result.sort(reverse=True)
+        return [elem for (elem, order) in result]
+
+    # General case, slowest method
+    it = iter(iterable)
+    result = [(key(elem), i, elem) for i, elem in zip(range(0, -n, -1), it)]
+    if not result:
+        return result
+    heapify(result)
+    top = result[0][0]
+    order = -n
+    _heapreplace = heapreplace
+    for elem in it:
+        k = key(elem)
+        if top < k:
+            _heapreplace(result, (k, order, elem))
+            top, _order, _elem = result[0]
+            order -= 1
+    result.sort(reverse=True)
+    return [elem for (k, order, elem) in result]
+
+# If available, use C implementation
+try:
+    from _heapq import *
+except ImportError:
+    pass
+try:
+    from _heapq import _heapreplace_max
+except ImportError:
+    pass
+try:
+    from _heapq import _heapify_max
+except ImportError:
+    pass
+try:
+    from _heapq import _heappop_max
+except ImportError:
+    pass
+
+
+if __name__ == "__main__":
+
+    import doctest # pragma: no cover
+    print(doctest.testmod()) # pragma: no cover