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 std::cmp::*;
use std::io::*;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();

    let mut event = Vec::new();
    let mut started = 0;
    let mut ended = 0;
    if n % 2 == 0 {
        event.push((1, 0));
        event.push((1, 1));
    }
    for _ in 0..n {
        let l: usize = itr.next().unwrap().parse().unwrap();
        let r: usize = itr.next().unwrap().parse().unwrap();
        event.push((l, 0));
        event.push((r + 1, 1));
    }
    event.sort();
    event.reverse();

    let mut ans = 0;
    let mut prev = 0;
    while let Some((pos, t)) = event.pop() {
        // prev ~ posは同じ状態
        if started > ended + n / 2 {
            if n % 2 == 0 {
                ans += pos - prev + 1;
            } else {
                ans += pos - prev;
            }
        }
        prev = pos;
        if t == 0 {
            // started - ended > n / 2の値は、中央値になることが可能。
            started += 1;
        } else {
            ended += 1;
        }
    }
    println!("{}", ans);
}