Implement and apply PEP 322, reverse iteration
diff --git a/Lib/heapq.py b/Lib/heapq.py
index 2c30b12..2223a97 100644
--- a/Lib/heapq.py
+++ b/Lib/heapq.py
@@ -165,7 +165,7 @@
# 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 xrange(n//2 - 1, -1, -1):
+ for i in reversed(xrange(n//2)):
_siftup(x, i)
# 'heap' is a heap at all indices >= startpos, except possibly for pos. pos