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 rep(i,cc,n) for(int i=cc;i<=n;++i)
#define drep(i,cc,n) for(int i=cc;i>=n;--i)
#define sz(s) (int)(s.size())
#define vecprint(v) rep(i,0,v.size()-1)cout << v[i] << " ";cout << endl;
#define arrprint(v,n) rep(i,0,n-1)cout << v[i] << " ";cout << endl;
#define matprint(v,n,m) rep(i,0,n-1){rep(j,0,m-1){cout << v[i][j] << " ";}cout << endl;}
#define SORT(v) sort(v.begin(), v.end())
#define REV(v) reverse(v.begin(), v.end())
using namespace std;
const int mod=1000000007;
const int INF=1001001001;
typedef long long ll;
typedef pair<int,int> P;
template<typename T>
using priority_queue_rev = priority_queue<T, vector<T>, greater<T>>; //小さいものから取り出す

int main(){
    int n;
    cin >> n;
    vector<int> a(n),b(n);
    rep(i,0,n-1)cin >> a[i] >> b[i];
    SORT(a);SORT(b);
    ll ans;
    if(n%2 == 1){
        ans = b[n/2] - a[n/2] + 1;
    }else{
        ll l = b[n/2-1] - a[n/2-1] + 1;
        ll r = b[n/2] - a[n/2] + 1;
        ans = l + r - 1;
    }


    cout << ans << endl;
//  printf("%.10f\n",ans);

    return 0;
}