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;

bool compare(vector<int> &l, vector<int> &r){
    return l.at(1) < r.at(1);
}

int main() {
    int N;
    cin >> N;
    vector<vector<int>> X(N, vector<int>(2));
    for(int i = 0; i < N; i++) {
        cin >> X.at(i).at(0);
        X.at(i).at(1) = i;
    }
    sort(X.begin(), X.end());
    vector<int> index_map(N);
    for(int i=0; i<N; i++){
        index_map.at(X.at(i).at(1)) = i;
    }

    int mid = (N - 1) / 2;
    for(int i = 0; i < N; i++) {
        int position = index_map.at(i);
        int m = (mid >= position) ? mid + 1 : mid;
        cout << X.at(m).at(0) << endl;
    }

    return 0;
}