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: Accepted)

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 3e3 + 5, mod = 998244353;


int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int n;
    cin >> n;
    vector<int> v, v2;
    for(int i = 0; i < n; i++){
        int x, y;
        cin >> x >> y;
        v.push_back(x);
        v2.push_back(y);
    }
    sort(v.begin(), v.end());
    sort(v2.begin(), v2.end());
    if(n&1) return cout << v2[n / 2] - v[n / 2] + 1, 0;
    int idx = n / 2;
    cout << v2[idx] + v2[idx - 1] + 1 - v[idx] - v[idx - 1];


    return 0;

}