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 proconio::{input, fastout};

#[fastout]
fn main() {
    input! {
        n: usize,
        ab: [(usize, usize); n],
    }

    let count = 
        if n % 2 == 0 {
            let a_m = (ab[n/2].0 + ab[n/2-1].0) as f64 / 2.;
            let b_m = (ab[n/2].1 + ab[n/2-1].1) as f64 / 2.;
            println!("{} {}", a_m, b_m);
            ((b_m - a_m) * 2.) as usize + 1
        } else {
            let a_m = ab[n/2].0;
            let b_m = ab[n/2].1;
            b_m - a_m + 1
        };

    println!("{}", count);
}