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)

use std::cmp::*;
use std::collections::*;
use std::io::*;
use std::str::*;

use std::string::*;

struct IO<R, W: std::io::Write>(R, std::io::BufWriter<W>);

impl<R: std::io::Read, W: std::io::Write> IO<R, W> {
    pub fn new(r: R, w: W) -> Self {
        Self(r, std::io::BufWriter::new(w))
    }

    pub fn write<S: ToString>(&mut self, s: S) {
        use std::io::Write;
        self.1.write(s.to_string().as_bytes()).unwrap();
    }

    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let s: String = self
            .0
            .by_ref()
            .bytes()
            .map(|b| b.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect();

        return s.parse::<T>().ok().unwrap();
    }

    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }

    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}

fn prime_factorization(n: i64) -> HashMap<i64, i64> {
    let mut n = n;
    let mut res = HashMap::new();
    let mut i = 2;
    while i * i <= n {
        while n % i == 0 {
            n /= i;
            let count = res.entry(i).or_insert(0);
            *count += 1;
        }
        i += 1;
    }

    if n > 1 {
        res.insert(n, 1);
    }

    return res;
}

fn main() {
    let (r, w) = (std::io::stdin(), std::io::stdout());
    let mut sc = IO::new(r.lock(), w.lock());
    let n = sc.read();
    let mut a = Vec::new();
    let mut b = Vec::new();
    for _ in 0..n {
        let aa: i64 = sc.read();
        let bb: i64 = sc.read();
        a.push(aa);
        b.push(bb);
    }

    a.sort();
    b.sort();

    if n % 2 == 1 {
        let mid_a = a[n / 2];
        let mid_b = b[n / 2];
        println!("{}", mid_b - mid_a + 1);
    } else {
        let mid_a = a[n / 2 - 1] + a[n / 2];
        let mid_b = b[n / 2 - 1] + b[n / 2];
        println!("{}", mid_b - mid_a + 1);
    }
}