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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <list>
#include <set>
#include <numeric>
#include <map>
#include <math.h>
#include <iomanip>
#include <stack>
#include <queue>
#include <bitset>
#include <math.h>

#define INF 100100100

typedef long long int llint;

using namespace std;

typedef pair<llint, llint>Pii;
typedef pair<double, double>Pdd;

#define pi 3.141592653589793)
#define mod 1000000007

int main() {
	int n;
	cin >> n;
	vector<int>x(n);
	for (int i = 0; i < n; i++)cin >> x[i];
	vector<int>a = x;
	sort(a.begin(), a.end());
	for (int i = 0; i < n; i++) {
		int res;
		if (a[n / 2 - 1] < x[i])res = a[n / 2 - 1];
		else if (a[n / 2 - 1] == x[i])res = a[n / 2];
		else if (a[n / 2] > x[i])res = a[n / 2];
		else if (a[n / 2] == x[i])res = a[n / 2 - 1];
		cout << res << endl;
	}
	return 0;
}