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: Runtime Error)

fn main() {
    proconio::input! {
        n: usize,
        abs: [(i64, i64); n],
    }
    let mut m = std::collections::BTreeMap::new();
    enum SE {
        Start,
        End,
    }
    for (a, b) in abs {
        m.entry(a).or_insert((0, SE::Start)).0 += 1;
        m.entry(b + 1).or_insert((0, SE::End)).0 += 1;
    }
    if n % 2 == 0 {
        let mid = (n / 2 - 1, n / 2);
        let mut mid_left = (0, 0);
        let mut mid_right = (0, 0);
        let mut start_count = 0;
        let mut end_count = 0;
        for (key, (value, se)) in m {
            match se {
                SE::Start => {
                    if (start_count..start_count + value).contains(&mid.0) {
                        mid_left.0 = key;
                    }
                    if (start_count..start_count + value).contains(&mid.1) {
                        mid_right.0 = key;
                    }
                    start_count += value;
                }
                SE::End => {
                    if (end_count..end_count + value).contains(&mid.0) {
                        mid_left.1 = key;
                    }
                    if (end_count..end_count + value).contains(&mid.1) {
                        mid_right.1 = key;
                    }
                    end_count += value;
                }
            }
        }
        let ans = ((mid_right.1 + mid_left.1) - (mid_right.0 + mid_left.0)) / 2 + 1;
        panic!();
        println!("{}", ans);
    } else {
        let mid = n / 2;
        let mut mid_range = (0, 0);
        let mut start_count = 0;
        let mut end_count = 0;
        for (key, (value, se)) in m {
            match se {
                SE::Start => {
                    if (start_count..start_count + value).contains(&mid) {
                        mid_range.0 = key;
                    }
                    start_count += value;
                }
                SE::End => {
                    if (end_count..end_count + value).contains(&mid) {
                        mid_range.1 = key;
                    }
                    end_count += value;
                }
            }
        }
        let ans = mid_range.1 - mid_range.0;
        println!("{}", ans);
    }
}