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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<iostream>
using namespace std;
const int N = 200005;
int a[N], b[N];
int main() {
    int n;
    cin >> n;
    for(int i = 1; i <= n; ++i) {
        cin >> a[i] >> b[i];
    }

    if(!(n&1)) {
        if(a[n/2] < a[n/2+1]) {
            if(b[n/2] + a[n/2+1] > b[n/2+1] + a[n/2]) {
                cout << (b[n/2]-a[n/2])*(b[n/2+1]-a[n/2+1]) << '\n';
            } else {
                cout << (b[n/2+1]-a[n/2]+1) << '\n';
            }
        } else {
            if(b[n/2] + a[n/2+1] > b[n/2+1] + a[n/2]) {
                cout << (b[n/2]-a[n/2])*(b[n/2+1]-a[n/2+1]) << '\n';
            } else {
                cout << (b[n/2+1]-a[n/2]+1) << '\n';
            }
        }
    } else {
        cout << b[n/2+1]-a[n/2+1]+1 << '\n';
    }
}