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 superslice::Ext;
fn main() {
    proconio::input! {
        n: usize,
        ab: [(i32,i32);n],
    }
    let mut a: Vec<_>=ab.iter().map(|&(a,_)|a).collect();
    a.sort();
    let mut b: Vec<_>=ab.iter().map(|&(_,b)|b).collect();
    b.sort();
    if n&1==1 {
        let (s,e) = f(n/2, n, &a, &b);
        println!("{}",e-s+1);
    }
    else {
        let (s1,e1) = f(n/2-1, n, &a, &b);
        let (s2,e2) = f(n/2, n, &a, &b);
        println!("{}",(e1+e2)-(s1+s2)+1)
    }
}
fn f(k: usize, n: usize, a: &Vec<i32>, b: &Vec<i32>) -> (i32,i32) {
    let mut l=a[0]-1;
    let mut r=b[n-1]+1;
    while l+1<r {
        let m=(l+r)/2;
        let (x,y)=(b.lower_bound(&m),n-a.upper_bound(&m));
        let z=n-x-y;
        if x+z<=k {
            l=m;
        }
        else {
            r=m;
        }
    }
    let s=r;
    let mut l=a[0]-1;
    let mut r=b[n-1]+1;
    while l+1<r {
        let m=(l+r)/2;
        let (x,y)=(b.lower_bound(&m),n-a.upper_bound(&m));
        let z=n-x-y;
        if y+z<=n-1-k {
            r=m;
        }
        else {
            l=m;
        }
    }
    let e=l;
    (s,e)
}