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)

fn read<T: std::str::FromStr>(
    stdin_lock: &mut std::io::StdinLock,
    buf: &mut Vec<u8>,
    delimiter: u8,
) -> T {
    buf.clear();
    let l = std::io::BufRead::read_until(stdin_lock, delimiter, buf).unwrap();
    buf.truncate(l - 1); // remove delimiter
    let s = unsafe { std::str::from_utf8_unchecked(&buf) };
    s.parse().unwrap_or_else(|_| panic!("read"))
}

fn main() {
    let stdin = std::io::stdin();
    let mut stdin_lock = stdin.lock();
    let mut buf: Vec<u8> = Vec::new();
    let n: usize = read(&mut stdin_lock, &mut buf, b'\n');
    let mut dv = vec![0i32; n];
    for i in 0..n - 1 {
        dv[i] = read(&mut stdin_lock, &mut buf, b' ');
    }
    dv[n - 1] = read(&mut stdin_lock, &mut buf, b'\n');

    dv.sort();

    let ml = dv[dv.len() / 2 - 1];
    let mr = dv[dv.len() / 2];
    let ans = if ml == mr { 0 } else { mr - ml };
    println!("{}", ans);
}