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 (1.42.0) (judge verdict: Accepted)

#![allow(unused, non_snake_case, dead_code, non_upper_case_globals)]
use {
    proconio::{marker::*, *},
    std::{cmp::*, collections::*, mem::*},
};

#[fastout]
fn main() {
    input! {//
        n:usize,
       mut d:[i64;n],
    }
    d.sort();
    println!("{}", d[n / 2] - d[n / 2 - 1]);
}
pub trait BinarySearch<T> {
    fn lower_bound_by<F: Fn(&T) -> bool>(&self, f: F) -> usize;
    fn lower_bound(&self, x: &T) -> usize;
    fn upper_bound(&self, x: &T) -> usize;
}
impl<T: Ord> BinarySearch<T> for [T] {
    fn lower_bound_by<F: Fn(&T) -> bool>(&self, f: F) -> usize {
        let mut ng = -1;
        let mut ok = self.len() as i64;
        while (ok as i32 - ng as i32).abs() > 1 {
            let mid = (ok + ng) / 2;
            if f(&self[mid as usize]) {
                ok = mid;
            } else {
                ng = mid;
            }
        }
        ok as usize
    }
    fn lower_bound(&self, x: &T) -> usize {
        self.lower_bound_by(|y| y >= x)
    }
    fn upper_bound(&self, x: &T) -> usize {
        self.lower_bound_by(|y| y > x)
    }
}