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)

//  Copyright © 2017 Adán López Alatorre. All rights reserved.
//

#include <iostream>
#include <vector>
#include <string.h>
#include <deque>
#include <queue>
#include <algorithm>
#include <set>
#include <map>
#include <sstream>
#include <stack>
#include <iomanip>
#include <climits>
#include <cmath>
#define fi first
#define se second
#define th second

using namespace std;

typedef long long ll;
typedef pair<int, int> ii;
typedef pair<ii, int> iii;
typedef pair<double, int> di;
typedef vector<int> vi;
typedef multiset<ll, greater<ll>>::iterator iit;
typedef pair<iit, iit> pvi;

const int maxN = 5e5 + 3, maxV = 1e5 + 5, LIM = 2e4;

int n, res[maxN];
ii arr[maxN];

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    
    cin >> n;
    
    for(int i = 0; i < n; i++) cin >> arr[i].fi, arr[i].se = i;
    
    sort(arr, arr + n);
    
    for(int i = 0; i < (n >> 1); i++){
        res[arr[i].se] = arr[n >> 1].fi;
    }
    for(int i = (n >> 1); i < n; i++){
        res[arr[i].se] = arr[(n >> 1) - 1].fi;
    }
    for(int i = 0; i < n; i++) cout << res[i] << '\n';
    
}