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>
#define itn int
#define REP(i,n) for(int i=0;i<(n);i++)
#define IREP(i,n) for(int i=n-1;i>=0;i--)
#define FOR(i,k,n) for(int i=(k);i<(n);i++)
#define REPEACH(itr,k) for(auto&& itr : k)
#define MOD 1000000007

using namespace std;
typedef long long ll;

const ll INF = 1LL<<60;
typedef pair<double,int> P;


int main(){
  int n;
  cin >> n;
  vector<pair<int,int>> v(n);
  REP(i,n){
    int a;
    cin >> a;
    v[i].first=a;
    v[i].second=i;
  }
  sort(v.begin(),v.end());
  vector<int> ans(n);
  REP(i,n){
    if(i<n/2){
      ans[v[i].second]=v[n/2].first;
    }
    else{
      ans[v[i].second]=v[n/2-1].first;
    }
  }
  REP(i,n){
    cout << ans[i] << endl;
  }
}