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)

pub struct Poper<'a> {
	iter: std::str::SplitWhitespace<'a>,
}

impl<'a> Poper<'a> {
	pub fn new(iter: std::str::SplitWhitespace) -> Poper {
		Poper { iter: iter }
	}
	pub fn read<T: std::str::FromStr>(&mut self) -> T
	where
		<T as std::str::FromStr>::Err: std::fmt::Debug,
	{
		self.iter.next().unwrap().parse().unwrap()
	}
}

fn main() {
	let input = {
		use std::io::Read;
		let mut s = String::new();
		std::io::stdin().read_to_string(&mut s).unwrap();
		s
	};
	let mut iter = Poper::new(input.split_whitespace());
	let n: usize = iter.read();

	let nums = {
		let mut t: Vec<u32> = (0..n).into_iter().map(|_| iter.read()).collect();
		t.sort();
		t
	};

	let len = nums.len();
	if 1 == len % 2 {
		println!("0");
		return;
	}

	let midi2 = len / 2;
	let midi1 = midi2 - 1;

	let mid1 = nums[midi1];
	let mid2 = nums[midi2];

	println!("{}", mid2 - mid1);
}