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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Scanner;

class Main{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int nyuryoku=sc.nextInt();
		int[] sort=new int[nyuryoku];
		for(int i=0; i<nyuryoku; i++) {
			sort[i]=sc.nextInt();
		}
		int[] seikei=new int[nyuryoku];
		for(int i=0; i<nyuryoku; i++) {
			seikei[i]=sort[i];
		}
		int[] copuy=new int[nyuryoku];
		copuy=seikei;

		Arrays.sort(copuy);

		int a=copuy[nyuryoku/2-1];
		int b=copuy[nyuryoku/2];
		for(int i=0; i<nyuryoku; i++) {
			if(sort[i]>a) {
				System.out.println(a);
			}
			else {
				System.out.println(b);
			}
		}
	}
}