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)

    //For higher stack size use g++ -O2 -std=c++11 -Wall -Wl,--stack=268435456 a.cpp -o a.exe in cmd
    #include <bits/stdc++.h>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <cstring>
    #include <chrono>
    #include <complex>
    #define ll long long
    #define ld long double
    #define vi vector<int>
    #define vll vector<ll>
    #define vvi vector < vi >
    #define pii pair<int,int>
    #define pll pair<long long, long long>
    #define mod 1000000007
    #define inf 1000000000000000001;
    #define all(c) c.begin(),c.end()
    #define mp(x,y) make_pair(x,y)
    #define mem(a,val) memset(a,val,sizeof(a))
    #define eb emplace_back
    #define pb push_back
    #define f first
    #define s second
    
    using namespace std;


    int main()
    {
        std::ios::sync_with_stdio(false);
        int T;
        T=1;
        // cin.ignore(); must be there when using getline(cin, s)
        while(T--)
        {
            ll n,i,j,a,b;
            cin>>n;
            ld x=0,z;
            vector<ll> v2,v;
            for(i=1;i<=n;++i)
            {
                cin>>a>>b;
                v2.pb(a);
                v.pb(b);
            }
            sort(v2.begin(),v2.end());
            sort(v.begin(),v.end());
            ld ans=-1,ans2=-1;
            if(n%2==1)
            {
                ans=v2[(n+1)/2-1];
                ans2=v[(n+1)/2-1];
            }
            else
            {
                ans=1.0*(v2[n/2-1]+v2[n/2])/2;
                ans2=1.0*(v[n/2-1]+v[n/2])/2;
            }
            if(n%2==1)
            {
                ll ans3=ans2-ans+1;
                cout<<ans3<<endl;
            }
            else
            {
                ll ans3 = (ans2-ans)*2+1;
                cout<<ans3<<endl;
            }
        }
        return 0;
    }