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)

macro_rules! get {
    (Vec<$t:ty>) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.split_whitespace()
                .map(|t| t.parse::<$t>().unwrap())
                .collect::<Vec<_>>()
        }
    };
    ($t:ty) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.trim().parse::<$t>().unwrap()
        }
    };
    ($($t:ty),*) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            (
                $(iter.next().unwrap().parse::<$t>().unwrap(),)*
            )
        }
    };
    ($t:ty; $n:expr) => {
        (0..$n).map(|_|
            get!($t)
        ).collect::<Vec<_>>()
    };
    ($($t:ty),*; $n:expr) => {
        (0..$n).map(|_|
            get!($($t),*)
        ).collect::<Vec<_>>()
    };
}

fn main() {
    let N = get!(usize);
    let mut d = get!(Vec<usize>);

    // dの右中央値と左中央値の差が解
    d.sort();
    let ans = d[N / 2] - d[N / 2 - 1];

    println!("{}", ans);
}