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>

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

int main(){

    int n,m,x[200001];
    int y[200001];
    int i,j,k,t;

    scanf("%d", &n);

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

    qsort(x, n, sizeof(int), asc);
    
    for(i=0;i<n-1;i++){
        for(j=n-1;j>i;j--){
            if(x[j-1]>x[j]){
                t = x[j];
                x[j] = x[j-1];
                x[j-1] = t;
            }
        }
    }

    m = n/2-1;

    for(i=0;i<n;i++){
        if(x[m]>=y[i]){
            printf("%d\n", x[m+1]);
        }else{
            printf("%d\n", x[m]);
        }
    }


    return 0;
}