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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <unordered_map>
#include <deque>
#include <list>
#include <set>
#include <random>

using namespace std;
using ll = long long;
const int BIG = 100000007;

using range = struct{
    ll min,max;
};

int main(){
    ll n;
    cin >> n;
    vector<range> veca(n,{0,0});
    vector<range> vecb(n,{0,0});
    for(int i = 0;i < n;++i){
        ll a,b;
        cin >> a >> b;
        veca[i].min = a;
        veca[i].max = b;
        vecb[i].min = a;
        vecb[i].max = b;
    }

    sort(veca.begin(),veca.end(),[](const auto& left, const auto& right){
        return left.min < right.min;
    });

    sort(vecb.begin(),vecb.end(),[](const auto& left, const auto& right){
        return left.max < right.max;
    });


    if(n % 2 == 1){
        cout << (vecb[n / 2].max - veca[n / 2].min + 1) << endl;
    }
    else{
        cout << ((vecb[n / 2 - 1].max + veca[n / 2].max) - (vecb[n / 2 - 1].min + veca[n / 2].min) + 1) << endl;
    }
}