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)

#pragma warning(disable : 4996)
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
// https://atcoder.jp/contests/abc169/tasks/abc169_e
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    ll ans = 0;
    ll n;
    cin >> n;
    vector<ll> a(n);
    vector<ll> b(n);
    for(int i = 0;i<n;i++){
        cin >> a[i] >> b[i];
    }
    sort(all(a));
    sort(all(b));
    if(n%2 == 0){
        ll ax = b[n/2] + b[(n/2)-1];
        ll in = a[n/2] + a[(n/2)-1];
        ans = ax - in + 1;
    }
    else{
        ll ax = b[n/2];
        ll in = a[n/2];
        ans = ax - in + 1;
    }
    cout << ans << "\n";

    return 0;
}