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)

use std::cmp::{max, min};
use std::error::Error;
use std::io;

pub fn main() {
    if let Err(_e) = run() {
        panic!("failed.");
    }
}

fn run() -> Result<(), Box<dyn Error>> {
    let mut input = String::new();
    io::stdin().read_line(&mut input)?;
    let n: i32 = input.trim().parse()?;
    input.clear();
    let mut ranges = vec![];
    let mut l_sort = vec![];
    let mut r_sort = vec![];
    for i in 0..n {
        io::stdin().read_line(&mut input)?;
        let a_b: Vec<i64> = input
            .trim()
            .split_whitespace()
            .map(|s| s.parse::<i64>().unwrap())
            .collect();
        ranges.push(a_b.clone());
        l_sort.push(a_b[0]);
        r_sort.push(a_b[1]);
        input.clear();
    }
    l_sort.sort_by(|e1, e2| e1.cmp(&e2));
    r_sort.sort_by(|e1, e2| e1.cmp(&e2));

    if n % 2 == 1 {
        let l = l_sort[n as usize / 2];
        let r = r_sort[n as usize / 2];
        println!("{}", r - l + 1);
    } else {
        let l = l_sort[n as usize / 2] + l_sort[n as usize / 2 - 1];
        let r = r_sort[n as usize / 2] + r_sort[n as usize / 2 - 1];
        println!("{}", r - l + 1);
    }

    Ok(())
}