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)

//! Framework <https://github.com/vain0x/procon>

#![allow(non_snake_case)]
#![allow(unused_imports)]

use std::collections::*;

pub struct Scan(Box<dyn Iterator<Item = &'static str>>); // '

impl Scan {
    fn new() -> Self {
        let mut buf = String::new();
        let read_line = move || {
            std::io::stdin().read_line(&mut buf).unwrap();
            Box::leak(buf.split_off(0).into_boxed_str()).split_whitespace()
        };
        Scan(Box::new(std::iter::repeat_with(read_line).flatten()))
    }

    pub fn word<T: std::str::FromStr>(&mut self) -> T {
        self.0.next().unwrap().parse().ok().unwrap()
    }

    pub fn list<T: std::str::FromStr>(&mut self, len: usize) -> Vec<T> {
        std::iter::repeat_with(|| self.word()).take(len).collect()
    }
}

fn median(x: &[i64]) -> i64 {
    let n = x.len();

    let mut x = x.to_owned();
    x.sort();

    if n % 2 == 0 {
        (x[n / 2 - 1] + x[n / 2]) / 2
    } else {
        x[n / 2]
    }
}

fn main() {
    let mut scan = Scan::new();

    let N = scan.word::<usize>();
    let mut A = vec![];
    let mut B = vec![];

    for _ in 0..N {
        let a = scan.word::<i64>();
        let b = scan.word::<i64>();

        let (a, b) = if N % 2 == 0 { (a * 2, b * 2) } else { (a, b) };

        A.push(a);
        B.push(b);
    }

    let am = median(&A);
    let bm = median(&B);

    let n = (bm - am + 1).max(0);

    println!("{}", n);
}