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>

int main(void)
{

  int i, j, tmp;

  /* 数値を格納する配列 */
  int number[200001];
  int a[200001];
  /* 数値の総数を入力 */
  int total;
  scanf("%d", &total);

  /* 配列に格納する数値を入力 */
  for (i=0; i<total; ++i)
    {scanf("%d", &number[i]);
     a[i]=number[i];}
  /* 数値を昇順にソート */
  for (i=0; i<total; ++i) {
    for (j=i+1; j<total; ++j) {
      if (number[i] > number[j]) {
        tmp =  number[i];
number[i] = number[j];
        number[j] = tmp;
      }
    }
  }

  /* 昇順ソートした数値を出力 */
  for(i=0;i<total;i++)
{if(a[i]<=number[total/2-1])
{printf("%d\n",number[total/2]);}
else
{printf("%d\n",number[total/2-1]);}
}
}