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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
using namespace std;

int fa[200001],numx[200001],numy[200001];

int fin(int x){
    if (fa[x]==x) return x;
    return fa[x]=fin(fa[x]);
}


int main()
{
    int n;scanf("%d",&n);
    for (int i=1;i<=2e5;++i) fa[i]=i;

    for (int i=0;i<n;++i){
        int x,y;scanf("%d%d",&x,&y);
        fa[fin(x) ]=fin(y+1e5);
    }

    for (int i=1;i<=1e5;++i) ++numx[fin(i)];
    for (int i=1e5+1;i<=2e5;++i) ++numy[fin(i)];

    //for (int i=1;i<=2*n;++i) printf("%d ",fin(i));

    long long ans=0;
    for (int i=1;i<=2e5;++i) ans+=1LL*numx[i]*numy[i];
    ans-=n;
    printf("%lld\n",ans);
}



