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::io::Read;


fn input() -> (usize,Vec<i64>,Vec<i64>) {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    
    let n:usize = iter.next().unwrap().parse().unwrap();
    let (a,b) = (0..n).map(|_|{
            let a:i64 = iter.next().unwrap().parse().unwrap();
            let b:i64 = iter.next().unwrap().parse().unwrap();
            (a,b)
        }).unzip();
    (n,a,b)
}

fn main() {
    let (n,mut a,mut b) = input();

    if n%2==0 {
        a.iter_mut().for_each(|v|{*v*=2});
        b.iter_mut().for_each(|v|{*v*=2});
    }


    a.sort();
    b.sort();
    let med_a = (a[n/2]+a[(n-1)/2])/2;
    let med_b = (b[n/2]+b[(n-1)/2])/2;
    let res = med_b-med_a+1;
    println!("{}",res);
}