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<stdio.h>
long compare(const void *a,const void *b){
  return *(long*)a-*(long*)b;
}

int main(){
  long n = 0;
  long a[200000] = {};
  long sorteda[200000] = {};
  long lowermid = 0;
  long highermid = 0;
  long i = 0;
  scanf("%ld",&n);
  for(i=0;i<n;i++){
    scanf("%lu",&a[i]);
  }
  memcpy(sorteda,a,sizeof(long)*n);
  qsort(sorteda,n,sizeof(long),compare);
  /*for(i=0;i<n;i++){
    printf("%ld\n",sorteda[i]);
  }*/
  lowermid = sorteda[n/2-1];
  highermid = sorteda[n/2];
  for(i=0;i<n;i++){
    if(a[i] >= highermid){
      printf("%d\n",lowermid);
    }else{
      printf("%d\n",highermid);
    }
  }
  return 0;
}