Problem:
Score : 500 points

Problem Statement

There are N integers X_1, X_2, \cdots, X_N, and we know that A_i \leq X_i \leq B_i.
Find the number of different values that the median of X_1, X_2, \cdots, X_N can take.

Notes

The median of X_1, X_2, \cdots, X_N is defined as follows. Let x_1, x_2, \cdots, x_N be the result of sorting X_1, X_2, \cdots, X_N in ascending order.

If N is odd, the median is x_{(N+1)/2};

if N is even, the median is (x_{N/2} + x_{N/2+1}) / 2.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq A_i \leq B_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 B_1
A_2 B_2
:
A_N B_N

Output

Print the answer.

Sample Input 1

2
1 2
2 3

Sample Output 1

3

If X_1 = 1 and X_2 = 2, the median is \frac{3}{2};

if X_1 = 1 and X_2 = 3, the median is 2;

if X_1 = 2 and X_2 = 2, the median is 2;

if X_1 = 2 and X_2 = 3, the median is \frac{5}{2}.

Thus, the median can take three values: \frac{3}{2}, 2, and \frac{5}{2}.

Sample Input 2

3
100 100
10 10000
1 1000000000

Sample Output 2

9991

Solution in Rust (1.42.0) (judge verdict: Accepted)

fn main() {
	let mut s = String::new();
	use std::io::Read;
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let mut a = vec![];
	let mut b = vec![];
	for _ in 0..n {
		let ai: i32 = s.next().unwrap().parse().unwrap();
		let bi: i32 = s.next().unwrap().parse().unwrap();
		a.push(ai);
		b.push(bi);
	}
	a.sort();
	b.sort();
	let (l, r) = if n % 2 == 0 {
		(a[n / 2 - 1] + a[n / 2], b[n / 2 - 1] + b[n / 2])
	} else {
		(a[n / 2], b[n / 2])
	};
	println!("{}", r - l + 1);
}