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: Runtime Error)

#include<stdio.h>
int main()
{
    int n;
    scanf("%d", &n);
    int x[20000];
    int i;
    for (i = 0; i < n; i++)
        scanf("%d", &x[i]);
    int y[20000];
    int max, min;
    int j;
    y[0] = x[0];
    for (i = 1; i < n; i++)
    {
        max = i;
        min = 0;
        if (x[i] < y[0])
        {
            for (j = i; j > 0; j--)
                y[j] = y[j - 1];
            y[0] = x[i];
            continue;
        }
        if (x[i] > y[max - 1])
        {
            y[i] = x[i];
            continue;
        }
        while (max - min > 1)
        {
            if (y[min] == x[i])
            {
                max = min;
                break;
            }
            if (y[max] == x[i])
            {
                min = max;
                break;
            }
            if (x[i] < y[(max + min) / 2])
                max = (max + min) / 2;
            else
                min = (max + min) / 2;
        }
        for (j = i; j > max; j--)
            y[j] = y[j - 1];
        y[max] = x[i];
    }
    int m1, m2;
    m1 = y[(n - 1) / 2];
    m2 = y[n / 2];
    for (i = 0; i < n; i++)
    {
        if (x[i] < m2)
            printf("%d\n", m2);
        else
            printf("%d\n", m1);
    }
    return 0;
}