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: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <iomanip>
#include <map>
#define rep(i,n) Rep(i,0,n)
#define Rep(i,k,n) for(int i=k ; i<n ; i++)
#define rep1(i,n) for(int i=1 ; i<=n ; i++)
#define vi vector<int>
#define vii vector<int,int>
#define Sort(v) sort(v.begin(),v.end())
#define Reverse(v) reverse(v.begin(),v.end())

//const int MOD = 1000000007;
//const int INF = 1<<30;
using namespace std;

string yn(bool x){
    return x ? "Yes" : "No";
}

int n;

void median(vi a,int i){
    a.erase(a.begin()+i);
    
    Sort(a);
    
    cout << a[(n/2)-1] << endl;
}

int main(){
    
    cin >> n;
    vi x(n);
    
    rep(i,n) cin >> x[i];
    
    rep(i,n){
        median(x,i);
    }
}