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 proconio::{
   fastout, input,
   marker::{Bytes, Chars, Isize1, Usize1},
};

fn main() {
   input! {
      n:usize
   }
   let mut vmax = vec![];
   let mut vmin = vec![];
   for _ in 0..n {
      input! {a:usize,b:usize}
      vmax.push(b);
      vmin.push(a);
   }
   vmax.sort();
   vmin.sort();
   if n % 2 == 0 {
      let max1 = vmax[n / 2 - 1];
      let max2 = vmax[n / 2];
      let min1 = vmin[n / 2 - 1];
      let min2 = vmin[n / 2];
      let mmax = (max2 + max1);
      let mmin = (min2 + min1);
      //dbg![max1, max2, min1, min2, mmax, mmin];
      println!("{}", mmax - mmin + 1);
   } else {
      let max = vmax[n / 2];
      let min = vmin[n / 2];
      println!("{}", max - min + 1);
   }
}