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;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::f64::consts::*;

#[allow(unused)]
const INF: usize = std::usize::MAX / 4;
#[allow(unused)]
const M: usize = 1000000007;

fn main() {
    input! {
        n: usize,
        ab: [(usize, usize); n],
    }
    let mut l = 0;
    let mut r = 10000000000;
    while r - l > 1 {
        let m = (l + r) / 2;
        let mut left = 0;
        for i in 0..n {
            if ab[i].0 < m {
                left += 1;
            }
        }
        if 2 * left < n {
            l = m;
        } else {
            r = m;
        }
    }
    let m_min = l;
    let mut l = 0;
    let mut r = 10000000000;
    while r - l > 1 {
        let m = (l + r) / 2;
        let mut right = 0;
        for i in 0..n {
            if m < ab[i].1 {
                right += 1;
            }
        }
        if 2 * right < n {
            r = m;
        } else {
            l = m;
        }
    }
    let m_max = r;
    eprintln!("{} {}", m_min, m_max);
    if n % 2 == 0 {
        println!("{}", 2 * (m_max - m_min) - 1);
    } else {
        println!("{}", m_max + 1 - m_min)
    }
}