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>
using namespace std;
#define ll long long

const ll nax=1e18;

bool is_prime(ll n)
{
 ll i;
    if(n==1
       ) return false;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0) return false;
    }
    return true;
}


void solve()
{
  ll n,i;
    cin>>n;
    ll ans=0;
   ll a[n][2];
    for(i=0;i<n;i++)
    {
        cin>>a[i][0]>>a[i][1];
    }
    if(n&1)
    {
       ans=a[n/2][1]-a[n/2][0]+1;
        cout<<ans<<endl;
        return ;
    }
    else
    {
     ll i=(n/2)-1;
     ll j=i+1;
      map<ll,ll>mp;
        mp[a[i][0]+a[j][0]]++;
        mp[a[i][0]+a[j][1]]++;
        mp[a[i][1]+a[j][0]]++;
        mp[a[i][1]+a[j][1]]++;
        cout<<mp.size()<<endl;
    }
}

int main()
{
 int t=1;
  //  cin>>t; 
    
    while(t--)
     solve();   
    
}