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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
void MergeSort(long long int x[ ],long long int left,long long int right);
        long long int temp[200001];
int main (void){
        int ans;
        int N;
        long long int A[200001];

        scanf("%d",&N);
        for (int i = 0; i < N; i++) {
                scanf("%lld",&A[i]);
        }

MergeSort(A, 0, N-1);
if(N==2){
  printf("%lld\n",A[1]);
  printf("%lld\n",A[0]);
}else{
for (int i = 0; i < N/2; i++) {
        printf("%lld\n",A[N/2]);
}
for (int i = 0; i < N/2; i++) {
        printf("%lld\n",A[(N/2)-1]);
}
}
        return 0;
}

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

    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--];
}