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 9.2.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
typedef long long ll;

int main(void)
{
  int n;
  scanf("%d",&n);
  ll x[200010];
  ll y[200010];
  for(int i=0;i<n;i++){
    scanf("%lld",&x[i]);
  }
  for(int i=0;i<n;i++){
    y[i]=x[i];
  }

  for(int i=n-1;i>0;i--){
    for(int j=i;j>=0;j--){
      if(y[i]<y[j]){
        ll tmp=y[i];
        y[i]=y[j];
        y[j]=tmp;
      }
    }
  }

  // for(int i=0;i<n;i++){
  //   printf("%lld ",y[i]);
  // }
  // puts("");

  for(int i=0;i<n;i++){
    if(x[i]<=y[n/2-1]){
      printf("%lld\n",y[n/2]);
    }else{
      printf("%lld\n",y[n/2-1]);
    }
  }

  return 0;
}