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>

struct number {
  int s, t;
};

int asc(const void *a, const void *b){
  struct number *A = (struct number *)a;
  struct number *B = (struct number *)b;
  return A->s - B->s;
}

int
main(int argc, char *argv[])
{
  int n; struct number x[200000];
  scanf("%d", &n);
  for(int i = 0; i < n; ++i){
    scanf("%d", &x[i].s);
    x[i].t = i;
  }

  qsort(x, n, sizeof(struct number), asc);

  int b[200000];
  for(int i = 0; i < n; i++){
    if(i <= (n / 2) - 1) b[x[i].t] = x[n / 2].s;
    else b[x[i].t] = x[n / 2 - 1].s;
  }

  for(int i = 0; i < n; i++) printf("%d\n", b[i]);

  return 0;
}