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>
typedef long long LL;
const int N=100000;
const int L=200000;
int n,x,y;
int fa[L+5];
int cntx[L+5],cnty[L+5];
LL ans;
int Find(int x) {
	if(fa[x]==x) return x;
	return fa[x]=Find(fa[x]);
}
int main() {
	for(int i=1;i<=L;i++)
		fa[i]=i;
	scanf("%d",&n);
	for(int i=1;i<=n;i++) {
		scanf("%d %d",&x,&y);
		int fx=Find(x),fy=Find(y+N);
		if(fx!=fy) fa[fy]=fx;
	}
	for(int i=1;i<=L;i++) {
		int fi=Find(i);
		if(i<=N) cntx[fi]++;
		else cnty[fi]++;
	}
	for(int i=1;i<=L;i++)
		ans+=(LL)cntx[i]*cnty[i];
	printf("%lld\n",ans-n);
	return 0;
}