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 <bits/stdc++.h>
using namespace std;
#define rep(i,n) for (int i = 0; i < (int)(n); ++i)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)


int main() {
  int n;
  cin >> n;
  vector<long>a(n),b(n);
  rep(i,n){
    cin>>a.at(i)>>b.at(i);
  }
  sort(a.begin(),a.end());
  sort(b.begin(),b.end());
  long ans;
  if(n%2==1){
    ans=b.at(n/2)-a.at(n/2)+1;
  }
  else{
    ans=((b.at(n/2-1)+b.at(n/2))/2-(a.at(n/2-1)+a.at(n/2))/2)*2+1;
  }
  
    cout<<ans<<endl;
  
}