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>

using namespace std;

#define ll long long
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define pii pair<ll,ll>
#define vpii vector<pair<ll,ll> >
#define F first
#define S second
#define ld long double
#define built __builtin_popcountll
#define mst(a,i) memset(a,i,sizeof(a))
#define all(x) x.begin(),x.end()
#define itit(it,a) for(auto it=(a).begin(); it!=(a).end(); it++)
#define rep(i,a,b) for(ll i=a; i<b; i++)
#define repr(i,a,b) for(ll i=a; i>b; i--)
#define reprr(i,a,b) for(ll i=a; i>=b; i--)
#define pi 3.14159265358979323846264338327950288419716939937510582097494459230
ll max3(ll x,ll y,ll z) {return max(max(x,y),z);}
ll min3(ll x,ll y,ll z) {return min(min(x,y),z);}
const ll M=2e5+10,M2=1e6+10,mod=1e9+7,inf=1e17+10;

int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	ll n;
	cin >> n;
	ll x[n];
	rep(i,0,n) cin >> x[i];

	ll y[n];
	rep(i,0,n) y[i]=x[i];

	sort(y,y+n);
	reverse(y,y+n);

	ll z=y[n/2-1];

	rep(i,0,n){
		if(x[i]<z) cout << z << "\n";
		else cout << y[n/2] << "\n";
	}
	return 0;
}
/* The judge is never wrong! Your code is buggy
   Look for:
 * * Read the problem carefully.
 * * Don't forget to sort(), mod, ll!!!!
 * * Initial value = +/- infinity instead of zero!!!
 * * an easier and alternate approach
 * * read the problem statement carefully
 * * if concept is correct and still WA, try with a different implementation
 * * special cases (n=1?)
 * * overflow (ll vs int?)
 * * array bounds
 * * if you have no idea just guess the appropriate well-known algorithm instead of doing nothing :/
 */