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: Time Limit Exceeded)

#include <stdio.h>
int main(){
	int n,i,j,temp,bmax,bmin;
	scanf("%d",&n);
	int a[n],b[n];
	for (i = 0;i < n;i++){
		scanf("%d",&a[i]);
	}
	for(i = 0;i<n;i++){
		b[i] = a[i];
	}
	for(i = 0;i < n/2;i++){
		bmax =i;
		bmin =i;
		for(j = i;j < n - i;j++){
			if(b[j] > b[bmax]) bmax = j;
			if(b[j] < b[bmin]) bmin = j;
		}
		temp = b[bmax];
		b[bmax] = b[n - 1 - i];
		b[n-1-i] =temp; /*最大値を交換*/
		temp = b[bmin];
		b[bmin] = b[i];
		b[i] =temp; /*最小値を交換*/
	}
	for(i = 0;i<n;i++){
		if(a[i] <= b[n/2-1]) printf("%d\n",b[n/2]);
		else printf("%d\n",b[n/2- 1] );
	}
}