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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define ll long long 
#define endl "\n"
#define pb push_back
#define f(i,n) for(i=0;i<n;i++)
#define F(i,a,b) for(i=a;a<=b;i++)
#define arr(a,n) for( i=0;i<n;i++)cin>>a[i];
#define fi first
#define se second
#define mp make_pair
#define mod 1000000007
#define YES cout<<"YES"<<endl;
#define Yes cout<<"Yes"<<endl;
#define NO cout<<"NO"<<endl;
#define No cout<<"No"<<endl;
#define yes cout<<"yes"<<endl;
#define no cout<<"no"<<endl;
#define vi vector<ll>
#define ed end()
#define bg begin()
#define sz size()
#define ln length()
#define s() sort(a,a+n);
#define sr() sort(a,a+n,greater<ll>());
#define v()  sort(v.begin(),v.end());
#define vr() sort(v.begin(),v.end(),greater<ll>());
#define mod 1000000007
#define fast() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
ll gcd(ll a, ll b){if(!b)return a;return gcd(b, a % b);}
ll power(ll x,ll y,ll p){ll res=1;x%=p;while(y>0){if(y&1)res=(res*x)%p;y=y>>1;x=(x*x)%p;}return res;}
int main() {
    /*#ifndef ONLINE_JUDGE
        // for getting input from input.txt
        freopen("input.txt", "r", stdin);
        // for writing output to output.txt
        freopen("output.txt", "w", stdout);
    #endif*/
        

fast();
//ll t;cin>>t;while(t--)
{
   ll n;
   cin>>n;
   ll a[n],b[n];
   for(int i=0;i<n;i++)
    {
      cin>>a[i];
      b[i]=a[i];
    }
    sort(b,b+n,greater<ll>());
    ll f=(n+1)/2-1;
    ll s=f+1;
    ll x=b[f],y=b[s];
    for(int i=0;i<n;i++)
    {
      if(a[i]==x)
        cout<<y<<endl;
      else
        cout<<x<<endl;
    }


}
return 0;
}