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>
#include <stdlib.h>

typedef struct{
  int num1;
  int num2;
}Int;

int Cmp(const void *a, const void *b){
  return ((Int *)a)->num2 - ((Int *)b)->num2;
}

void Medians(int N, Int X[N]){
  int i, j;
  qsort(X, N, sizeof(Int), Cmp);
  for(i = 0;i < N;i++){
    for(j = 0;j < N / 2;j++){
      if(X[j].num1 == i){
	printf("%d\n", X[N/2].num2);
	break;
      }
    }
    if(j == N / 2) printf("%d\n", X[N/2-1].num2);
  }
}

void Do(int N){
  int i;
  Int X[N];
  for(i = 0;i < N;i++){
    X[i].num1 = i;
    scanf("%d", &X[i].num2);
  }
  Medians(N, X);
}

int main(void){
  int N;
  scanf("%d", &N);
  Do(N);
  return 0;
}