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 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>

#define rep(i, a, n)  for(int i=a; i<n; i++)
#define per(i, a, n)  for(int i=n-1; i>=a; i--)
#define fill0(n)  setfill('0') << right << setw(n)
#define all(x)  (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define LONGMAX 1e18
#define INTMAX 1e9
using namespace std;
typedef long long ll;

int n;
vector <ll> vx;
ll x[200100];

int main(){
  cin >> n;
  rep(i,1,n+1){
    cin >> x[i];
    vx.push_back(x[i]);
  }
  sort(all(vx));
  ll med1 = vx[n/2-1];
  ll med2 = vx[n/2];
  rep(i,1,n+1){
    int hoge = lower_bound(all(vx), x[i])-vx.begin();
    if(hoge <= n/2)std::cout << med2 << '\n';
    else std::cout << med1 << '\n';
    // if(x[i] > med1)std::cout << med1 << '\n';
    // else std::cout << med2 << '\n';
  }
}