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;
vector<long long> a,b;
int main()
{
	int n,i,j,tmp;
	long long best;
	scanf("%d",&n);
	a.resize(n);
	b.resize(n);
	for(i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
		b[i]=a[i];
	}
	sort(a.begin(),a.end());
	best=a[n/2-1];
	for(i=0;i<n;i++)
	{
		if(b[i]==a[n-1])
			tmp=n-1;
		else
			tmp=upper_bound(a.begin(),a.end(),b[i])-a.begin();
		if(n/2==tmp&&n/2-1!=i)
			printf("%lld\n",best);
		else
			printf("%lld\n",a[tmp]);
		//printf("%d\n",a[n/2-1]);
	}
	return 0;
}