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

#define swap(type, x, y)  do {type t = x; x = y; y = t;} while (0)

void bubble(int a[], int n)
{
    int k = 0;
    while (k < n - 1) {
        int j;
        int last = n - 1;
        for (j = n - 1; j > k; --j) {
            if (a[j - 1] > a[j]) {
                swap(int, a[j - 1], a[j]);
                last = j;
            }
        }
        k = last;
    }
}

void quick(int a[], int left, int right)
{
    int pl = left;
    int pr = right;
    int x = a[(pl + pr) / 2];
    do {
        while (a[pl] < x) ++pl;
        while (a[pr] > x) --pr;
        if (pl <= pr) {
            swap(int, a[pl], a[pr]);
            ++pl;
            --pr;
        }
    } while (pl <= pr);
    if (left < pr) quick(a, left, pr);
    if (pl < right) quick(a, pl, right);
}

int main(void)
{
    int n, i, j;
    scanf("%d", &n);

    int *x, *y;
    x = calloc(n, sizeof(int));
    y = calloc(n, sizeof(int));
    for (i = 0; i < n; ++i) {
        scanf("%d", x + i);
        y[i] = x[i];
    }

    quick(y, 0, n - 1);
    int ans1 = y[n / 2 - 1];
    int ans2 = y[n / 2];

    for (i = 0; i < n; ++i) {
        if (x[i] <= ans1) printf("%d\n", ans2);
        else printf("%d\n", ans1);
    }
    return 0;
}