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>
#include <string.h>
#include <math.h>
 
int compare_int(const void *a, const void *b)
{
    return *(long long*)a - *(long long*)b;
}

int min_comp(int min, int a)
{

}
int max_comp(int max, int a)
{
  if(max>a)    return max;
  else    return a;
}


int main()
{
  int N;
  long long X[200001]={};
  long long Y[200001]={};
  long long temp = 0;
  long long mid0 = 0;
  long long mid1 = 0;
  int sel = 0;
  int cost_N = 0;
  int number;
  int debug_cnt;
//  long long N[101]={};
  int i,j;
  int cont;
//    int k=0;

  scanf("%d", &N);

  for(i=0;i<N;i++){
    scanf("%lld", &X[i]);
    Y[i] = X[i];
  }
  
  qsort(X, N, sizeof(long long), compare_int); 
 //   for(j=0;j<N;j++){
 //     printf("%lld  ", X[j]);
 //   }
  mid0 = X[(N-1)/2];
  mid1 = X[((N-1)/2)+1];
 //  printf(" %d %lld %lld\n", (N-1)/2,mid0,mid1);
  for(i=0;i<N;i++){
    if(mid0<Y[i]){
      printf("%lld\n", mid0);
    } else {
      printf("%lld\n", mid1);
    }
  }
//  X_mid = (X_sum - X[0]);
  //debug
//  printf("%lld\n", (X_mid/2));
  
  //  else printf("No\n"); 
return 0;
}