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>
#define int long long
#define mp make_pair
#define dd second
#define ff first
#define N 200005
#define pb push_back
#define pp pair<int,int>
using namespace std;

int n;
int tab[N];

int tab2[N];

main() {
    ios_base::sync_with_stdio(false);
    cin >> n;
    for(int i = 0; i < n; ++i) cin >> tab[i], tab2[i] = tab[i];

    sort(tab2, tab2+n);

    for(int i = 0; i < n; ++i) {
        int val = tab[i];
        int index = lower_bound(tab2, tab2+n, val)- tab2;
        int k = (n-1)/2;
        if(k < index) {
            cout << tab2[k] << " ";
        } else {
            cout << tab2[k+1] << " ";
        }
    }
    cout << endl;

}