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++14 (GCC 5.4.1) (judge verdict: Accepted)

    #include <bits/stdc++.h>
    using namespace std;
     
    int main() {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
     
        long n;
        cin >> n;
     
        long x[n];
        long x_copy[n];
     
        for (long i = 0; i < n; i++) {
            cin >> x[i];
            x_copy[i] = x[i];
        }
        sort(x_copy, x_copy + n);
        long small_mean = x_copy[n / 2 - 1];
        long large_mean = x_copy[n / 2 ];
     
        for (long i = 0; i < n; i ++) {
            if (x[i] <= small_mean) {
                cout << large_mean << "\n";
            } else {
                cout << small_mean << "\n";
            }
        }
     
        return 0;
    }