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};

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;

fn check(must: usize, can: usize, n: usize) -> bool {
    if n % 2 == 1 {
        let m = (n + 1) / 2;
        return must < m && m <= can;
    } else {
        let m1 = n / 2;
        let m2 = n / 2 + 1;
        return (must < m1 && m1 <= can) || (must < m2 && m2 <= can);
    }
}

#[fastout]
fn main() {
    input! {
        n: usize,
        ab: [(usize, usize); n],
    }

    let mut abh = BinaryHeap::new();
    let mut bbh = BinaryHeap::new();
    for (a, b) in &ab {
        abh.push(Reverse(*a));
        bbh.push(Reverse(*b + 1));
    }

    let mut ans = 0;
    let mut can = 0;
    let mut must = 0;
    let mut pre = 0;
    let dummy = 2e9 as usize;
    while !abh.is_empty() || !bbh.is_empty() {
        let Reverse(a) = abh.pop().unwrap_or(Reverse(dummy));
        let Reverse(b) = bbh.pop().unwrap_or(Reverse(dummy));

        if a < b {
            if check(must, can, n) {
                ans += a - pre;
            }
            pre = a;
            can += 1;
            if b != dummy {
                bbh.push(Reverse(b));
            }
        } else {
            if check(must, can, n) {
                ans += b - pre;
            }
            pre = b;
            must += 1;
            if a != dummy {
                abh.push(Reverse(a));
            }
        }
    }
    println!("{}", ans);
}