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)

/*
 ============================================================================
 Name        : AtCoder_C.c
 Author      : 
 Version     :
 Copyright   : Your copyright notice
 Description : Hello World in C, Ansi-style
 ============================================================================
 */

#include <stdio.h>
#include <stdlib.h>

int N;
void MergeSort(int x[ ], int left, int right)
{
    int mid, i, j, k;
    int temp[N];

    if (left >= right)              /* 配列の要素がひとつなら */
        return;                     /* 何もしないで戻る */

                                    /* ここでは分割しているだけ */
    mid = (left + right) / 2;       /* 中央の値より */
    MergeSort(x, left, mid);        /* 左を再帰呼び出し */
    MergeSort(x, mid + 1, right);   /* 右を再帰呼び出し */

      /* x[left] から x[mid] を作業領域にコピー */
    for (i = left; i <= mid; i++)
        temp[i] = x[i];

      /* x[mid + 1] から x[right] は逆順にコピー */
    for (i = mid + 1, j = right; i <= right; i++, j--)
        temp[i] = x[j];

    i = left;         /* i とj は作業領域のデーターを */
    j = right;        /* k は配列の要素を指している */

    for (k = left; k <= right; k++)    /* 小さい方から配列に戻す */
        if (temp[i] <= temp[j])        /* ここでソートされる */
            x[k] = temp[i++];
        else
            x[k] = temp[j--];
}
int main(void) {
	int i;

	  /* 数値を格納する配列 */
	int total;
		  scanf("%d", &total);
	  int* number = (int*)malloc(sizeof(int)*total);
	  int* sortmae = (int*)malloc(sizeof(int)*total);

	  /* 数値の総数を入力 */


	  /* 配列に格納する数値を入力 */
	  for (i=0; i<total; ++i){
	    scanf("%d", &number[i]);
	    sortmae[i] = number[i];
	  }
	  N = total;
	  MergeSort(number, 0, total -1);

	  for (i=0; i<total; ++i){
		  if(sortmae[i] <= number[(total/2-1)] ){
			  printf("%d\n", number[total/2 ]);
		  }else{
			  printf("%d\n", number[total/2-1]);
		  }
	  }
	return EXIT_SUCCESS;
}