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 ll = long long;
using namespace std;
const int INFint = 1e9+1;
const ll INFll = (ll)1e18+1;
ll MOD=1e9+7;


int main(){
  int N;
  cin>>N;
  vector<int> A(N),B(N);
  for(int i(0);i<N;i++){
    cin>>A[i]>>B[i];
  }
  sort(A.begin(),A.end());
  sort(B.begin(),B.end());
  if(N % 2 == 1){
    int hidari = A[N/2];
    int migi = B[N/2];
    cout << migi - hidari + 1 << endl;
  }else{
    int hidari = (A[N/2]+A[N/2])/2;
    int migi = (B[N/2]+B[N/2])/2;
    cout << (migi - hidari)*2+1 << endl;
  }
  return 0;
}