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 <string.h>
#include <stdlib.h>

void split(int out[], char str[]);
int cmp(const void *p, const void *q);

int main(void) {
  int nums;
  char *tmp;
  int *list, *sorted;
  int rmed, lmed;
  tmp = (char *)malloc(sizeof(char) * 2000001);
  list = (int *)malloc(sizeof(int) * 200001);
  sorted = (int *)malloc(sizeof(int) * 200001);
  int i;
  scanf("%d\n", &nums);
  gets(tmp);
  split(list, tmp);
  memcpy(sorted, list, sizeof(int) * nums);
  qsort(sorted, nums, sizeof(int), cmp); // listをソート
  rmed = sorted[nums / 2];
  lmed = sorted[nums / 2 - 1]; // 正しい
  free(sorted);
  for (i = 0; i < nums; ++i) {
    if (list[i] <= lmed) {
      printf("%d\n", rmed);
    } else {
      printf("%d\n", lmed);
    }
  }
  free(tmp);
  free(list);
  return 0;
}

void split(int out[], char str[]) {
  int i;
  int out_i = 0, tmp_i = 0;
  char tmp[11] = {};
  for (i = 0; ; ++i) {
    if (str[i] == ' ') {
      out[out_i++] = atoi(tmp);
      memset(tmp, '\0', sizeof(tmp));
      tmp_i = 0;
    } else if (str[i] == '\0') {
      out[out_i] = atoi(tmp);
      break;
    } else {
      tmp[tmp_i++] = str[i];
    }
  }
}

int cmp(const void *p, const void *q) {
    return *(int*)p - *(int*)q;
}