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)

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().split_whitespace()
        .map(|e| e.parse().ok().unwrap()).collect()
}
fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
    let mut v2 = Vec::new();
    for _ in 0..n {
        let mut s = String::new();
        std::io::stdin().read_line(&mut s).ok();
        let v = s.trim().split_whitespace()
            .map(|e| e.parse().ok().unwrap()).collect();
        v2.push(v);
    }
    v2
}
fn get_half_count(len: u32, nums:Vec<Vec<u64>>) -> u64 {
    let anums:Vec<u64> = nums.iter().map(|nums| nums[0]).collect();
    let bnums:Vec<u64> = nums.iter().map(|nums| nums[1]).collect();
  
    return getHalf(&bnums, len) - getHalf(&anums, len) + 1;
}

fn main() {
    let n:u32 = read();
    let arrays:Vec<Vec<u64>> = read_vec2(n);
    println!("{}", get_half_count(n, arrays));
}

fn getHalf(nums:&Vec<u64>, len:u32) -> u64 {
    let mut arrays = vec![0u64; len as usize];
    arrays.copy_from_slice(nums);
    arrays.sort();
    let way_point:usize = (len / 2) as usize;
    return if len % 2 == 1 { // 奇数
        arrays[way_point]
    } else { //偶数
        arrays[way_point] + nums[way_point - 1]
    }
}

#[test]
fn test_getHalf() {
    let vec = vec!(1, 10, 100);
    assert_eq!(getHalf(&vec, 3), 10);
}

#[test]
fn test_get_half_count() {
    let n = 3;
    let nums = vec!(
        vec!(100, 100),
        vec!(10, 10000),
        vec!(1, 1000000000),
    );
    assert_eq!(get_half_count(n, nums), 9991);
}