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;
typedef long long ll;
using P = pair<int,int>;
/* ちゃんと考えてわかって実装 */

int main(void){
    int n; cin >> n;
    vector<ll> A(n), B(n);
    for(int i=0; i<n; i++){
        cin >> A[i] >> B[i];
    }

    ll A_med, B_med, ans;
    if(n % 2 != 0){
        A_med = A[(n-1)/2];
        B_med = B[(n-1)/2];
        ans = B_med - A_med + 1LL;
        cout << ans << endl;
    }
    else if(n % 2 == 0){
        A_med = (A[n/2] + A[(n/2)-1]);
        B_med = (B[n/2] + B[(n/2)-1]);
        ans = B_med - A_med + 1LL;
        cout << ans << endl;
    }
    return 0;
}