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>

static int compare_ll (const void *a, const void *b) {
    long long *a_ = (long long*)a;
    long long *b_ = (long long*)b;
    if ( *a_ > *b_ )
        return 1;
    else if ( *a_ < *b_ )
        return -1;
    return 0;
}

int main (void) {
    int n;
    scanf("%d", &n);
    long long x[n], sorted[n];
    for ( int i=0; i<n; i++ ) {
        scanf("%lld", &x[i]);
        sorted[i] = x[i];
    }

    qsort(sorted, n, sizeof(long long), compare_ll);

    long long ans_cand_max = sorted[n / 2];
    long long ans_cand_min = sorted[n / 2 - 1];

    for ( int i=0; i<n; i++ ) {
        if ( x[i] >= ans_cand_max )
            printf("%lld\n", ans_cand_min);
        else
            printf("%lld\n", ans_cand_max);
    }

    return EXIT_SUCCESS;
}