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 <iostream>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
using namespace std;
using ll = long long;

int main(int argc, const char * argv[]) {
    int n;
    cin >> n;
    vector<pair<int, ll>> v(n);
    
    ll x;
    pair<ll, int> p;
    for (int i = 0; i < n; i++) {
        cin >> x;
        p .first = x;
        p.second = i;
        v[i] = p;
    }
    sort(v.begin(), v.end());
    int a[n];
    for (int i = 0; i < n; i++) {
        a[v[i].second] = i;
    }
    for (int i = 0; i < n; i++) {
        if (a[i] < n / 2) {
            cout << v[(n+1)/2+1-1].first << endl;
        }else{
            cout << v[(n+1)/2-1].first << endl;
        }
    }
    return 0;
}