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)

macro_rules! get_one {
    ($t:ty) => {{
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().parse::<$t>().unwrap()
    }};

    ($t:ty; $n:expr) => {
        (0..$n).map(|_| get_one!($t)).collect::<Vec<_>>()
    };
}

macro_rules! get_tuple {
    ($($t:ty),*) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            (
                $(iter.next().unwrap().parse::<$t>().unwrap(),)*
            )
        }
    };

    ($($t:ty),*; $n:expr) => {
        (0..$n).map(|_|
            get_tuple!($($t),*)
        ).collect::<Vec<_>>()
    };
}

fn main() {
    let n = get_one!(isize);
    let v_ab = get_tuple!(isize, isize; n);

    let mut v_a: Vec<_> = v_ab.iter().map(|&(a, b)| a).collect();
    let mut v_b: Vec<_> = v_ab.iter().map(|&(a, b)| b).collect();

    let ans = median(&mut v_b) - median(&mut v_a) + 1;

    println!("{}", ans)



}

fn median(numbers: &mut Vec<isize>) -> isize {

    numbers.sort();

    let mid = numbers.len() / 2;
    if numbers.len() % 2 == 0 {
        numbers[mid - 1] + numbers[mid]
    } else {
        numbers[mid]
    }

}