There are N dots in a two-dimensional plane. The coordinates of the i-th dot are (x_i, y_i).

We will repeat the following operation as long as possible:

* Choose four integers a, b, c, d (a \neq c, b \neq d) such that there are dots at exactly three of the positions (a, b), (a, d), (c, b) and (c, d), and add a dot at the remaining position.



We can prove that we can only do this operation a finite number of times. Find the maximum number of times we can do the operation.

Constraints

* 1 \leq N \leq 10^5
* 1 \leq x_i, y_i \leq 10^5
* If i \neq j, x_i \neq x_j or y_i \neq y_j.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
x_1 y_1
:
x_N y_N


Output

Print the maximum number of times we can do the operation.

Examples

Input

3
1 1
5 1
5 5


Output

1


Input

2
10 10
20 20


Output

0


Input

9
1 1
2 1
3 1
4 1
5 1
1 2
1 3
1 4
1 5


Output

16
#include<bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<=b;i++)
using namespace std;

typedef unsigned long long ULL;
typedef long long ll;
const ULL base = 233;
const int N = 100005;

int n;
int f[201000];
ll val1[200100],val2[201000];
int find(int x){return f[x]==x?x:(f[x]=find(f[x]));}

int main(){
    cin>>n;
    FOR(i,1,2*N)f[i]=i;
    FOR(i,1,n){
        int x,y;
        scanf("%d%d",&x,&y);
        f[find(x)]= find(y+N);
    }
    FOR(i,1,N)val1[find(i)]++;
    FOR(i,N+1,2*N)val2[find(i)]++;
    ll ans=0;
    FOR(i,1,2*N){
        ans+=val1[i]*val2[i];
        //if(val1[i]*val2[i])cout<<ans<<" "<<i<<" "<<i-N<<endl;
    }
    printf("%lld\n",ans-n);
}
