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 <cstdio>
#include <iostream>
#define N 100000
#define xx fucker
#define yy lover
using namespace std;
int n,fa[N*2+5],xx[2*N+5],yy[2*N+5];
int find(int x){
	if(x==fa[x])return x;
	return fa[x]=find(fa[x]);
}
void work(int x,int y){
	int fx=find(x),fy=find(y);
	if(fx==fy)return;
	fa[fy]=fx;
	xx[fx]+=xx[fy],yy[fx]+=yy[fy];
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=N;i++)fa[i]=i,xx[i]=1;
	for(int i=N+1;i<=2*N;i++)fa[i]=i,yy[i]=1;
	for(int i=1,x,y;i<=n;i++){
		scanf("%d%d",&x,&y);
		work(x,y+N);
	}
	long long ans=0;
	for(int i=0;i<=2*N;i++)if(fa[i]==i)
		ans+=1ll*xx[i]*yy[i];
	cout<<ans-n<<endl;
	return 0;
}