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 Ruby (judge verdict: Accepted)

class PriorityQueue
  #?????????
  def initialize
    @queue = Array.new
  end

  #???????????????????´??????????
  def top
    @queue[0]
  end

  #???????????????????´??????????????????????????´??????????
  def pop
    top = @queue[0]
    n = @queue.size - 1

  end

  #?????§?????´??°
  def max_heap_fy(i)
    l = 2 * i
    r = 2 * i + 1
    if l<= @queue.size && @queue[l] > @queue[i] then largest = l
    else largest = i end
    largest = r if r <= @queue.size && @queue[r] > @queue[largest]
    if largest != i then
      @queue[i],@queue[largest] = @queue[largest],@queue[i]
      max_heap_fy(largest)
    end
  end
end


def max_heap_fy(i)
  l = 2 * i
  r = 2 * i + 1
  if l<= @h && @que[l] > @que[i] then largest = l
  else largest = i end
  largest = r if r <= @h && @que[r] > @que[largest]
  if largest != i then
    @que[i],@que[largest] = @que[largest],@que[i]
    max_heap_fy(largest)
  end
end

@h = gets.to_i
@que = gets.split.map(&:to_i)
@que.insert(0,-114514)
(@h / 2).downto(1)do |i|
  max_heap_fy(i)
end
puts " #{@que[1..@que.size - 1].join(" ")}"