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;

fn main() {
    let mut input = String::new();
    io::stdin().read_line(&mut input).ok();
    let n = input.trim().parse().unwrap();

    let mut input = String::new();
    io::stdin().read_line(&mut input).ok();

    let mut v: Vec<usize> = Vec::new();
    for z in input.trim().split_whitespace() {
        v.push(z.parse().unwrap());
    }

    let mut v2: Vec<usize> = Vec::new();
    v.sort();

    for z in 0..n {
        let mut buf;
        if z == 0 {
            buf = v[z];
        } else {
            buf = v[z] - v[z - 1];
        }

        v2.push(buf);
    }

    println!("{}", v2[n / 2]);
}