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;
#[allow(unused_imports)]
use num::Integer;
#[allow(unused_imports)]
use proconio::derive_readable;
use proconio::fastout;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::cmp;

fn double_median(v: &mut Vec<i64>) -> i64 {
  v.sort();
  let n = v.len();
  if n % 2 == 0 {
    v[n / 2 - 1] + v[n / 2]
  } else {
    v[n / 2]
  }
}

#[fastout]
fn main() {
  input! {
    n: usize,
    mut ab: [(i64, i64); n],
  }
  let lb = double_median(&mut ab.iter().map(|&(a, _)| a).collect());
  let ub = double_median(&mut ab.iter().map(|&(_, b)| b).collect());

  println!("{}", ub - lb + 1);
}