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 JAVA (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;



public class Main {
	
	InputUtil in = new InputUtil();

	public static void main(String[] args) throws IOException {
		new Main().exec();
	}

	public void exec() throws IOException {
		PrintWriter out = new PrintWriter(System.out);
		int H = in.nextInt();
		int[] heapData = in.nextInt(new int[H+1],1,H);
		MaximumHeap mh = new MaximumHeap(heapData);
		out.print(' ');
		OutputUtil.output(out, mh.getAll(), 1, H, " ");
		out.flush();
	}

	
}

class BinaryHeap {
	protected int[] data = null;
	
	public BinaryHeap(int[] data) {
		this.data = data;
	}
	
	public int[] getAll() {
		return data;
	}

	public int getParent(int i){
		if(i<=1 || data.length<=i) return -1;
		return i/2;
	}

	public int getLeft(int i){
		int leftId = i*2;
		return (i<=0 || data.length<=leftId) ? -1 : leftId;
	}

	public int getRight(int i){
		int rightId = i*2 + 1;
		return (i<=0 || data.length<=rightId) ? -1 :rightId;
	}
}

class MaximumHeap extends BinaryHeap {	
	
	public MaximumHeap(int[] data) {
		super(data);
		this.buildMaxHeap();
	}
	
	protected void maxHeapify(int i) {
		int leftId = getLeft(i);
		int rightId = getRight(i);
		int largest = i;
		if(leftId != -1 && data[largest] < data[leftId]) largest = leftId;
		if(rightId != -1 && data[largest] < data[rightId]) largest = rightId;
		if(largest != i){
			data[0] = data[i]; data[i] = data[largest]; data[largest] = data[0];
			maxHeapify(largest);
		}
	}
	
	public void buildMaxHeap() {
		for(int i=(data.length-1)/2; i>=1; i--)
			maxHeapify(i);
			
	}
	
}


class OutputUtil {
	public static final String EOF = System.lineSeparator();

	public static void output(PrintWriter out, int[] a, int off, int len, String delim) {
		out.print(a[off]);
		for(int i=off+1; i<off+len; i++){
			out.print(delim);
			out.print(a[i]);
		}
		out.println("");
		out.flush();
	}
}


class InputUtil {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    public String[] readStrArray(String delim) throws NumberFormatException, IOException{
    	return br.readLine().split(delim);
    }
    
    public int nextInt() throws NumberFormatException, IOException{
    	return Integer.parseInt(br.readLine());
    }
    
    public int[] nextInt(int[] a, int off, int num) throws NumberFormatException, IOException{
        int i=off;
        while(i<off+num) {
            for(String s: br.readLine().split(" ")){
            	if(i<off+num) {
                    a[i++] = Integer.parseInt(s);
            	}
            }
        }
        return a;
    }
}