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: Runtime Error)

#include <stdio.h>

int N;
int T[500010] ={0};

void swap(int T[], int x, int y)
{
    int tmp = T[x];
    T[x] = T[y];
    T[y] = tmp;
}

void maxHeapify(int T[], int i)
{
    int parent = i;
    int left = (i+1)*2 - 1;
    int right = (i+1)*2;
    
    int largest;
    if (T[parent] < T[right] && T[left] < T[right]) {
        largest = right;
    }
    else if (T[parent] < T[left] && T[right] < T[left]) {
        largest = left;
    }
    else {
        largest = parent;
    }
    
    /*
    int j;
    printf("p:%d -- ", i);
    for (j = 0; j < 10; j++) {
        printf(" %d", T[j]);
    }puts("");
    */
    
    if (largest != parent) {
        swap(T, largest, parent);
        maxHeapify(T, largest);
    }    
    
    return;
}


int main()
{
    int i, j;
    scanf("%d", &N);
    for (i = 0; i < N; i++) {
        scanf("%d", &T[i]);
    }
    for (i = N/2+1; i >= 0; i--) {
        maxHeapify(T, i);
    }
    for (i = 0; i < N; i++) {
        printf(" %d", T[i]);
    }
    puts("");
 
    return 0;
}