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>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <limits>
#include <map>
#include <vector>
#include <queue> 
#define LL long long
#define ft first
#define sd second
#define mp(x,y) make_pair(x,y)
//#define int LL
#define db double
using namespace std;
const int N   = 2e5+233;
//const int mod = ;
const int INF =numeric_limits<int >::max();
#define For(n) for (int i=1;i<=n;++i)
#define rep(i,x,y) for (int i=x;i<=y;++i)
void read(int &x)
{
	x=0;
	char ch=getchar();
	int f=1;
	while (!isdigit(ch)) (ch=='-'?f=-1:0),ch=getchar();
	while ( isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	x*=f;
}

int n,a[N],pos[N],b[N],k;

signed main()
{
	read(n);
	rep(i,1,n) read(a[i]),b[i]=a[i];
	sort(b+1,b+1+n);
	k=n/2;
	if (b[k]==b[k+1])
	{
		rep(i,1,n) printf("%d\n",b[k]);
		return 0;
	}
	rep(i,1,n) if (a[i]<=b[k]) printf("%d\n",b[k+1]);else printf("%d\n",b[k]);
	return 0;
}