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)

struct Scanner {
    handle: std::io::Stdin,
    buffer: std::collections::VecDeque<String>,
}

impl Scanner {
    fn new() -> Scanner {
        Scanner {
            handle: std::io::stdin(),
            buffer: std::collections::VecDeque::new(),
        }
    }
    fn next<T: std::str::FromStr>(&mut self) -> Result<T, T::Err> {
        if self.buffer.len() == 0 {
            loop {
                let mut available = String::new();
                self.handle.read_line(&mut available).unwrap();
                self.buffer = available.split_whitespace().map(String::from).collect();
                if self.buffer.len() > 0 {
                    break;
                }
            }
        }
        T::from_str(&self.buffer.pop_front().unwrap())
    }
}

fn main() {
    let mut scanner = Scanner::new();
    let n: usize = scanner.next().unwrap();
    let mut av = Vec::<i32>::new();
    let mut bv = Vec::<i32>::new();
    for _ in 0..n {
        av.push(scanner.next().unwrap());
        bv.push(scanner.next().unwrap());
    }
    av.sort();
	bv.sort();
	#[cfg(debug_assertions)] dbg!(&av, &bv);
    let total;
    if n % 2 == 1 {
        let min = av[(n - 1) / 2];
        let max = bv[(n - 1) / 2];
        total = max - (min - 1);
    } else {
        let min = av[(n - 1) / 2] + av[n / 2];
        let max = bv[(n - 1) / 2] + bv[n / 2];
        total = max - (min - 1);
    }
    println!("{}", total);
}