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: Wrong Answer)

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

int sortInt(const void *a, const void *b)
{
	return *(int *)a - *(int *)b;
}

int main(void)
{
	int N;
	int *X, *nX;
	int i, j;
	int m1, m2;

	scanf("%d", &N);

	X = (int *)malloc(sizeof(int) * N);
	nX = (int *)malloc(sizeof(int) * N);

	for(i = 0; i < N; i++)
	{
		scanf("%d", &X[i]);
		nX[i] = X[i];
	}

	qsort(nX, N, sizeof(int), sortInt);

	m1 = nX[N / 2];
	m2 = nX[(N / 2) - 1];

	for(i = 0; i < N; i++)
	{
		if(X[i] > m2)
			printf("%d\n", m1);
		else
			printf("%d\n", m2);
	}

		return 0;
}