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};
// use proconio::marker::*;

#[fastout]
fn main() {
   input! {
      n: usize,
      mut ab: [(i64, i64); n]
   }
   let mut ba: Vec<_> = ab.iter().map(|&(a, b)| (-b, -a)).collect();
   ab.sort();
   ba.sort();
   if n % 2 == 1 {
      let x = (n - 1) / 2;
      let ans = -ba[x].0 - ab[x].0 + 1;
      println!("{}", ans);
   } else {
      let x1 = n / 2 - 1;
      let x2 = n / 2;
      let l1 = ab[x1].0;
      let r1 = -ba[x2].0;
      let l2 = ab[x2].0;
      let r2 = -ba[x1].0;
      let mut ans = 0;
      if r1 < l2 {
         ans += r2 - l2 + 1;
         ans += r1 - l1;
      } else {
         ans += r2 - r1 + 1;
         ans += (r1 - l2) * 2;
         ans += r1 - l1;
      }
      println!("{}", ans);
   }
}