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 C (judge verdict: Accepted)

/* Max heap */
#include <stdio.h>
#include <stdlib.h>

void maxHeapify(int heap[], int i);
void buildMaxHeap(int heap[]);
void swap(int *, int *);

int H;      /* ヒープサイズ */

int main()
{
  int *heap, i;

  /* ヒープ用の領域を動的に確保 */
  scanf("%d", &H);
  heap = (int *)malloc( sizeof(int) * (H+1) );

  /* ヒープの各節点の値を読込む */
  for( i = 1; i<=H; i++ )
    scanf("%d", &heap[i]);
  
  /* 最大ヒープを作成 */
  buildMaxHeap(heap);

  /* 最大ヒープの各節点の値を節点の番号が１からHに向って順番に出力 */
  for( i = 1; i<=H; i++ )
    printf(" %d", heap[i]);
  printf("\n");
  
  return 0;
}


/* 配列heapを最大ヒープに変換する関数 */
void buildMaxHeap(int heap[])
{
  int i;

  for( i = H/2; i>=1; i-- )
    maxHeapify(heap, i);
}


/* heap[i]を根ノードとする部分木を最大ヒープに変換する関数 */
void maxHeapify(int heap[], int i)
{
  int l = i*2, r = i*2+1, largest;

  /* 根ノードheap[i]、左の子、右の子の中から最大のものを求める */
  if( l<=H && heap[i] < heap[l] )
    largest = l;
  else
    largest = i;
  if( r<=H && heap[largest] < heap[r] )
    largest = r;

  if( i != largest )
    {
      swap(&heap[i], &heap[largest]);
      maxHeapify(heap, largest);
    }
}


/* int型変数 a,b を交換する関数 */
void swap(int *a, int *b)
{
  int tmp;

  tmp = *a;
  *a  = *b;
  *b  = tmp;
}