| """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 = 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 | 
 | an 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() | 
 |         if key is None: | 
 |             result = min(it, default=sentinel) | 
 |         else: | 
 |             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 = result[0][0] | 
 |                 order += 1 | 
 |         result.sort() | 
 |         return [r[0] for r 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 = result[0][0] | 
 |             order += 1 | 
 |     result.sort() | 
 |     return [r[2] for r 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() | 
 |         if key is None: | 
 |             result = max(it, default=sentinel) | 
 |         else: | 
 |             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 = result[0][0] | 
 |                 order -= 1 | 
 |         result.sort(reverse=True) | 
 |         return [r[0] for r 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 = result[0][0] | 
 |             order -= 1 | 
 |     result.sort(reverse=True) | 
 |     return [r[2] for r 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 | 
 |     print(doctest.testmod()) |