Problem:
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Maximum Heap

A binary heap which satisfies max-heap property is called max-heap. In a max-heap, for every node $i$ other than the root, $A[i] \leq A[parent(i)]$, that is, the value of a node is at most the value of its parent. The largest element in a max-heap is stored at the root, and the subtree rooted at a node contains values no larger than that contained at the node itself.

Here is an example of a max-heap.

Write a program which reads an array and constructs a max-heap from the array based on the following pseudo code.

$maxHeapify(A, i)$ move the value of $A[i]$ down to leaves to make a sub-tree of node $i$ a max-heap. Here, $H$ is the size of the heap.

1 maxHeapify(A, i)
2 l = left(i)
3 r = right(i)
4 // select the node which has the maximum value
5 if l ≤ H and A[l] > A[i]
6 largest = l
7 else
8 largest = i
9 if r ≤ H and A[r] > A[largest]
10 largest = r
11
12 if largest ≠ i　// value of children is larger than that of i
13 swap A[i] and A[largest]
14 maxHeapify(A, largest) // call recursively

The following procedure buildMaxHeap(A) makes $A$ a max-heap by performing maxHeapify in a bottom-up manner.

1 buildMaxHeap(A)
2 for i = H/2 downto 1
3 maxHeapify(A, i)

Input

In the first line, an integer $H$ is given. In the second line, $H$ integers which represent elements in the binary heap are given in order of node id (from $1$ to $H$).

Output

Print values of nodes in the max-heap in order of their id (from $1$ to $H$). Print a single space character before each value.

Constraint

$1 \leq H \leq 500,000$

$-2,000,000,000 \leq$ value of a node $\leq 2,000,000,000$

Sample Input 1

10
4 1 3 2 16 9 10 14 8 7

Sample Output 1

16 14 10 8 7 9 3 2 4 1

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in Python3 (judge verdict: Wrong Answer)

def heapify(values):
  n = len(values)
  for i in range(1,n):
    upheap(values,i)
    print(' '.join(map(str, values)))

def upheap(values, i):
  while parent(i) >= 0:
    if values[i] > values[parent(i)]:
      values[i], values[parent(i)] = values[parent(i)], values[i]
      i = parent(i)
    else:
      break

def pop(values):
  ret = values[0]
  values[0], values[len(values)-1] = values[len(values)-1], values[0]
  del values[len(values)-1]
  downheap(values)
  return ret

def downheap(values):
  i = 0
  while left(i) < len(values) or right(i) < len(values):
    l = left(i)
    r = right(i)
    if l >= len(values):
      m = r
    elif r >= len(values):
      m = l
    else:
      m = r
      if values[l] > values[r]:
        m = l
    if values[i] < values[m]:
      values[m], values[i] = values[i], values[m]
    i = m

def parent(i):
  i = i+1
  return i//2-1

def left(i):
  i = i+1
  return i*2-1

def right(i):
  i = i+1
  return i*2


n = int(input())
a = list(map(int,input().split(' ')))

heapify(a)
print(' '.join(map(str, a)))