Problem:
Score : 300 points

Problem Statement

Takahashi made N problems for competitive programming.
The problems are numbered 1 to N, and the difficulty of Problem i is represented as an integer d_i (the higher, the harder).

He is dividing the problems into two categories by choosing an integer K, as follows:

A problem with difficulty K or higher will be for ARCs.

A problem with difficulty lower than K will be for ABCs.

How many choices of the integer K make the number of problems for ARCs and the number of problems for ABCs the same?

Problem Statement

2 \leq N \leq 10^5

N is an even number.

1 \leq d_i \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
d_1 d_2 ... d_N

Output

Print the number of choices of the integer K that make the number of problems for ARCs and the number of problems for ABCs the same.

Sample Input 1

6
9 1 4 4 6 7

Sample Output 1

2

If we choose K=5 or 6, Problem 1, 5, and 6 will be for ARCs, Problem 2, 3, and 4 will be for ABCs, and the objective is achieved.
Thus, the answer is 2.

Sample Input 2

8
9 1 14 5 5 4 4 14

Sample Output 2

0

There may be no choice of the integer K that make the number of problems for ARCs and the number of problems for ABCs the same.

Sample Input 3

14
99592 10342 29105 78532 83018 11639 92015 77204 30914 21912 34519 80835 100000 1

Sample Output 3

42685

Solution in Rust (1.15.1) (judge verdict: Accepted)

use std::io;

macro_rules! get {
    ($t: ty) => {
        {
            let mut x = String::new();
            io::stdin().read_line(&mut x).ok();
            x = x.trim_right().to_owned();
            x.parse::<$t>().unwrap()
        }
    };
}

macro_rules! insert_vec {
    ($t: ty) => {
        {
            let mut x = String::new();
            io::stdin().read_line(&mut x).ok();
            x = x.trim_right().to_owned();
            x.split_whitespace().map(|x| x.parse::<$t>().unwrap())
                .collect::<Vec<_>>()
        }
    };
}

fn main() {
    let n = get!(usize);
    let mut p = insert_vec!(i32);
    let mut ans; 
    p.sort();
    if p[ n / 2 - 1] == p[ n /2 ] {
        println!("0");
    } else {
        ans = p[n/2] - p [n/2 -1];
        println!("{}", ans);
    };
}