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::BufRead;
use std::str::FromStr;

fn main() {
    let stdin = std::io::stdin();
    let mut t = Tokenizer::new(stdin.lock().lines().map(|x| x.unwrap()));

    let num: i32 = t.take();
    let mut nums: Vec<i32> = Vec::new();
    for _i in 0 .. num {
        nums.push( t.take() );
    }
    let result = problem( &mut nums);
    println!("{}", result);
}

fn problem( nums: &mut Vec<i32> ) -> i32 {
    nums.sort();
    let num = nums.len();
    if (num % 2) == 1 {
        // 奇数個の場合
        return 0;
    }
    nums[num/2] - nums[num/2 -1]
}


#[cfg(test)]
mod test {
    use super::*;
    #[test]
    fn test1() {
        let mut nums: Vec<i32> = vec![9, 1, 4, 4, 6, 7];
        let result = problem( &mut nums );
        assert_eq!(result,2);
    }

    #[test]
    fn test2() {
        let mut nums: Vec<i32> = vec![9, 1, 14, 5, 5, 4, 4, 14];
        let result = problem( &mut nums );
        assert_eq!(result,0);
    }

    #[test]
    fn test3() {
        let mut nums: Vec<i32> = vec![99592, 10342, 29105, 78532, 83018, 11639, 92015, 77204, 30914, 21912, 34519, 80835, 100000, 1];
        let result = problem( &mut nums );
        assert_eq!(result,42685);
    }
}


struct Tokenizer<S: Iterator<Item = String>> {
    input: S,
    items: Vec<String>,
}

impl<S: Iterator<Item = String>> Tokenizer<S> {
    pub fn new(input: S) -> Self {
        Tokenizer {
            input: input,
            items: Vec::new(),
        }
    }

    pub fn take<T: FromStr>(&mut self) -> T {
        loop {
            if let Some(r) = self.items.pop() {
                return r.parse().ok().expect("parse failed");
            }
            let mut items = self
                .input
                .next()
                .unwrap()
                .split_whitespace()
                .map(|x| x.to_string())
                .collect::<Vec<String>>();
            items.reverse();
            self.items = items;
        }
    }
}