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)

#[allow(unused_imports)]
use itertools::Itertools;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;

fn main() {
    input! {
        n: usize,
        mut ab: [(i64, i64); n],
    }
    ab.sort();
    let m1;
    if n % 2 == 1 {
        m1 = ab[n / 2].0 * 2;
    } else {
        m1 = ab[n / 2].0 + ab[n / 2 - 1].0;
    }

    ab.sort_by_key(|x| x.1);

    let m2;
    if n % 2 == 1 {
        m2 = ab[n / 2].1 * 2;
    } else {
        m2 = ab[n / 2].1 + ab[n / 2 - 1].1;
    }

    // println!("{} {}", m1, m2);

    if n % 2 == 1 {
        println!("{}", (m2 - m1 + 1) / 2 + 1);
    } else {
        println!("{}", m2 - m1 + 1);
    }
}