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;
using ll = long long;

#define DUMP(x) cout << #x << " = " << (x) << endl;
#define REP(i, n) for(size_t i = 0; i < (n); i++)
#define REPR(i, n) for(size_t i = (n); i >= 0; i--)
#define FOR(i, a, b) for(size_t i = (a); i < (b); i++)
#define FOREACH(x,a) for(auto& (x) : (a))

const int INF = 1e9;
const int MOD = 1e9 + 7;

int N;
vector<int> X;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> N;
    vector<int> sort_X;
    REP(i, N) {
        int x; cin >> x;
        X.push_back(x);
        sort_X.push_back(x);
    }
    sort(sort_X.begin(), sort_X.end());
    REP(i, N) {
        if(X[i] > sort_X[N / 2 - 1]) cout << sort_X[N / 2 - 1] << "\n";
        else cout << sort_X[N / 2] << "\n";
    }
}