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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

//c++ テンプレ
#include<bits/stdc++.h>
using namespace std;

typedef long long llint;
typedef long double ld;

#define inf 1e18
#define mod 1000000007
#define sort(v) sort(v.begin(),v.end())
#define reverse(v) reverse(v.begin(),v.end())

priority_queue<llint,vector<llint>,greater<llint> > que;
priority_queue<llint> Que;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }


void solve(){
    int n;
    cin >> n;
    vector<llint>x(n+1);
    for(int i=1;i<=n;i++){
        cin >> x[i];
    }
    sort(x);
    llint res1=x[n/2];
    llint res2=x[n/2+1];
    for(int i=1;i<=n;i++){
        if(x[i]>=res2){
            cout << res1 << endl;
        }
        else{
            cout << res2 << endl;
        }
    }
}

int main(){
    solve();
    return 0;
}