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>
#include <stdint.h>
#include <stdlib.h>
#include <memory.h>

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

int main(void)
{
    uint32_t N;
    uint32_t X[200000];
    uint32_t tmp[200000];

    /* read data */
    scanf("%u", &N);
    for (uint32_t i = 0; i < N; i++) {
        scanf("%u", &X[i]);
    }
    
    /* calc base index */
    const uint32_t baseIdx = N / 2;

    /* calc B_i */
    for (uint32_t i = 0; i < N; i++) {
        memcpy(tmp, X, sizeof(X));
        tmp[i] = 0;
        qsort(tmp, N, sizeof(uint32_t), cmp);
        printf("%u\n", tmp[baseIdx]);
    }
    return 0;
}