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::{fastout, input};

#[fastout]
fn main() {
    input! {in_n: u64, in_ab: [[u64; 2]; in_n]};
    let mut a = Vec::new();
    let mut b = Vec::new();

    for ab in &in_ab {
        a.push(ab[0]);
        b.push(ab[1]);
    }

    a.sort();
    b.sort();

    if a.len() % 2 == 0 {
        let i = a.len() / 2;
        let minval = a[i - 1] + b[i - 1];
        let maxval = a[i] + b[i];
        let v = maxval - minval + 1;

        println!("{}", v);
    } else {
        let i = a.len() / 2;
        let v = b[i] - a[i] + 1;

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