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: Wrong Answer)

use std::io;
use std::str::FromStr;

fn input<T:FromStr>() -> T {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.trim().parse().ok().unwrap()
}

fn input_tuple() -> (u64, u64) {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    let vec: Vec<u64> = line.split_whitespace().map(|s| s.parse().unwrap()).collect();
    (vec[0], vec[1])
}

fn main() {
    let n: usize = input();
    let mut a_vec: Vec<u64> = Vec::new();
    let mut b_vec : Vec<u64> = Vec::new();
    for _ in 0..n {
        let (a, b) = input_tuple();
        a_vec.push(a);
        b_vec.push(b);
    }
    a_vec.sort();
    b_vec.sort();
    // println!("{:?}", a_vec);
    // println!("{:?}", b_vec);
    if n % 2 == 1 {
        // println!("{}", a_vec[n / 2]);
        // println!("{}", b_vec[n / 2]);
        println!("{}", b_vec[n/2] - a_vec[n/2] + 1);
    } else {
        let low_lim_0 = a_vec[n / 2 - 1];
        let up_lim_0 = b_vec[n / 2 - 1];
        let low_lim_1 = a_vec[n - n / 2];
        let up_lim_1 = b_vec[n - n / 2];
        // if up_lim_0 <= low_lim_1 {
            println!("{}", up_lim_1 + low_lim_1 - (low_lim_1 + low_lim_0) + 1);
        // } else {

        // }
    }
}