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>

/* ソート関数 */
int int_sort( const void * a , const void * b ) {
  return *(const int *)a - *(const int *)b;
}

int main(void) {
    int N;
    scanf("%d", &N);

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

    qsort(X, N, sizeof(X[0]), int_sort);

    // 各中央値計算    
    int B[N];
    for(i=0;i<N;i++){
        int center_index = ( (N-1) + 1 ) / 2 - 1;
        if(i <= center_index) {
            B[i] = X[center_index+1];
        } else {
            B[i] = X[center_index];
        }
    }

    // 各中央値表示
    for(i=0;i<N;i++){
        printf("%d\n", B[i]);
    }

    return 0;
}