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 main() {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let n: usize = iter.next().unwrap().parse().unwrap();

    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    let mut ds = buf.split_whitespace()
        .map(|x| x.parse::<usize>().unwrap())
        .collect::<Vec<_>>();

    ds.sort();
    let ds_min = ds[0];
    let ds_max = ds.last().unwrap();

    let mut ans = 0;
    for k in ds_min..(ds_max+1) {
        let mut lb = -1;
        let mut ub = n as isize;
        while ub - lb > 1 {
            let mid = ((lb+ub)/2) as usize;
            if ds[mid] < k {
                lb = mid as isize;
            }
            else {
                ub = mid as isize;
            }
        }
        if lb == ((n/2-1) as isize) {
            ans += 1;
        }
    }

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