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>

#define rep(i, n) for(int i = 0; i < (n); ++i)

using namespace std;

using ll = long long;
using P = pair<int, int>;

int main(){
    int n; cin >> n;

    vector<int> v(n), s(n);
    rep(i, n) {
        cin >> v[i] >> s[i];
    }
    sort(v.begin(), v.end());
    sort(s.begin(), s.end());

    vector<int> num(4);
    num[0] = v[n-2];
    num[1] = v[n-1];
    num[2] = s[0];
    num[3] = s[1];

    ll ans = 0;
    if(num[0] != num[3]) {
        ans += abs(num[3]-num[0]);
    } else {
        ++ans;
    }

    if(num[1] != num[2]) {
        ans += abs(num[2] - num[1]);
    } else {
        ++ans;
    }
    cout << ans << endl;


}