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
/*Lucky_Glass*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=1e5;
int n;
int pre[2*N+5],cntX[2*N+5],cntY[2*N+5];
int Find(int u){return pre[u]=(pre[u]==u? u:Find(pre[u]));}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=2*N;i++) pre[i]=i;
	for(int i=1;i<=n;i++){
		int x,y;scanf("%d%d",&x,&y);
		int Fx=Find(x),Fy=Find(N+y);
		if(Fx!=Fy) pre[Fx]=Fy;
	}
	for(int i=1;i<=2*N;i++){
		int Fi=Find(i);
		if(i<=N) cntY[Fi]++;
		else cntX[Fi]++;
	}
	ll ans=0;
	for(int i=1;i<=2*N;i++)
		ans+=1ll*cntX[i]*cntY[i];
	printf("%lld\n",ans-n);
	return 0;
}