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;
use std::fmt::Debug;
use std::str::FromStr;
#[allow(dead_code)]
fn read_as_vec<T>() -> Vec<T>
where
    T: FromStr,
    <T as FromStr>::Err: Debug,
{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim()
        .split_whitespace()
        .map(|c| T::from_str(c).unwrap())
        .collect()
}

#[allow(dead_code)]
fn read_as_string() -> String {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn main() {
    let n: usize = read_as_vec()[0];
    let mut uppers: Vec<usize> = vec![];
    let mut lowers: Vec<usize> = vec![];
    for _ in 0..n {
        let line = read_as_vec::<usize>();
        uppers.push(line[1]);
        lowers.push(line[0]);
    }

    uppers.sort();
    lowers.sort();
    dbg!(&uppers);
    if n % 2 == 1 {
        println!("{}", uppers[(n - 1) / 2] - lowers[(n - 1) / 2] + 1);
    } else {
        println!(
            "{}",
            uppers[(n - 1) / 2] + uppers[(n - 1) / 2 + 1]
                - (lowers[(n - 1) / 2] + lowers[(n - 1) / 2 + 1])
                + 1
        );
    }
}