Problem:
Score : 300 points

Problem Statement

When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l.

You are given N numbers X_1, X_2, ..., X_N, where N is an even number.
For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i.

Find B_i for each i = 1, 2, ..., N.

Constraints

2 \leq N \leq 200000

N is even.

1 \leq X_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X_1 X_2 ... X_N

Output

Print N lines.
The i-th line should contain B_i.

Sample Input 1

4
2 4 4 3

Sample Output 1

4
3
3
4

Since the median of X_2, X_3, X_4 is 4, B_1 = 4.

Since the median of X_1, X_3, X_4 is 3, B_2 = 3.

Since the median of X_1, X_2, X_4 is 3, B_3 = 3.

Since the median of X_1, X_2, X_3 is 4, B_4 = 4.

Sample Input 2

2
1 2

Sample Output 2

2
1

Sample Input 3

6
5 5 4 4 3 3

Sample Output 3

4
4
4
4
4
4

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>

int comp( const void *c1, const void *c2 );

int main(void){
	int n;
	scanf("%d",&n);
	int a[n];
	int b[n];
	int i;
	for(i=0;i<n;i++){
		scanf("%d",&a[i]);
		b[i] = a[i];
	}
	qsort( a, n, sizeof(int), comp );//a[]はソート済み
	//b[i]を比較する。
	for(i=0;i<n;i++){
		if(b[i] >= a[n/2]){
			printf("%d\n",a[(n/2)-1]);
		}else{
			printf("%d\n",a[n/2]);
		}
	}
	return 0;
}


/* 比較関数 */
int comp( const void *c1, const void *c2 )
{
  int tmp1 = *(int *)c1;
  int tmp2 = *(int *)c2;

  if( tmp1 < tmp2 )  return -1;
  if( tmp1 == tmp2 ) return  0;
  if( tmp1 > tmp2 )  return  1;
}