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>
using namespace std;
#define N 100100
typedef long long ll;
int n,f[N<<1];
int getf(int x){return x==f[x]?x:f[x]=getf(f[x]);}
int sz1[N<<1],sz2[N<<1];

int main(){
    ll ans=0;
    cin>>n;
    for(int i=1;i<N+N;++i)f[i]=i;
    int u,v;
    for(int i=1;i<=n;++i)
    {
        cin>>u>>v;
        f[getf(u)]=getf(v+N);
    }
    for(int i=1;i<=N;++i)
        sz1[getf(i)]+=1;
    for(int i=N+1;i<N+N;++i)
        sz2[getf(i)]+=1;
    for(int i=1;i<N+N;++i)ans+=1ll*sz1[i]*sz2[i];
    cout<<ans-n<<endl;
    return 0;
}