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>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
    int N, minv, maxv;  cin >> N;
    long ans;
    std::vector<int> v(N), u(N);
    for(int i=0; i<N; i++){
        int A, B;   cin >> A >> B;
        v[i] = A;
        u[i] = B;
    }
    std::sort(v.begin(),v.end());
    std::sort(u.begin(),u.end());
    if(N%2 == 1){
        minv = v[int((N-1)/2)];
        maxv = u[int((N-1)/2)];
        ans = maxv - minv + 1;
    }else{
        minv = (v[int(N/2) - 1] + v[int(N/2) - 1]) / 2;
        maxv = (u[int(N/2) - 1] + u[int(N/2) - 1]) / 2;
        ans = (maxv - minv)*2 + 1;
    }
    cout << ans;
}