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 Rust (1.15.1) (judge verdict: Accepted)

use std::cmp::*;
use std::io::*;

fn main() {
    let mut s: String = String::new();
    stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let mut x = Vec::new();
    for i in 0..n {
        let a: usize = itr.next().unwrap().parse().unwrap();
        x.push((a, i));
    }
    x.sort();

    let mut ans = vec![0; n];
    let mut out = Vec::new();
    for i in 0..n {
        // (n-1) / 2
        // x[n / 2 + 1]について
        if i < n / 2 {
            ans[x[i].1] = x[n / 2].0;
        } else {
            ans[x[i].1] = x[n / 2 - 1].0;
        }
    }
    for i in 0..n {
        writeln!(out, "{}", ans[i]).ok();
    }
    stdout().write_all(&out).unwrap();
}