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: Time Limit Exceeded)

use std::io::*;
use std::str::FromStr;
use std::collections::HashMap;
use std::collections::BTreeSet;
use std::collections::HashSet;
use std::cmp;
use std::f64::consts;

fn main() {
    let n: i64 = read();
    let mut vec: Vec<i64> = (0..n).map(|_| read()).collect();
    let mut sorted_vec: Vec<i64> = vec.clone();
    sorted_vec.sort();

    let middle_index: usize = (n as usize) / 2 - 1;

    for &i in vec.iter() {
        let mut new_vec = sorted_vec.clone();
        let rm_index = new_vec.iter().position(|&r| r == i).unwrap();
        new_vec.remove(rm_index);

        println!("{}", new_vec[middle_index]);
    }
}

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

// 最大公約数
fn gcd(a: i32, b: i32) -> i32 {
    match b {
        0 => a,
        _ => gcd(b, a % b)
    }
}

// 最小公倍数
fn lcm(a: i32, b: i32) -> i32 {
    a * b / gcd(a, b)
}

// 階乗
fn kaijou(n: i32)->i32 {
    if n == 1 {
        return n;
    }
    return n * kaijou(n-1);
}