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<vector>
#include<algorithm>
using namespace std;

typedef long long ll;
const int O=1e5;

int fa[2*O+5];
ll cnt1[2*O+5],cnt2[2*O+5];

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

int main() {
	#ifdef LOACL
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
	#endif
	for(int i=0;i<=2*O;i++)
		fa[i]=i;
	int N;
	scanf("%d",&N);
	for(int i=1;i<=N;i++) {
		int x,y;
		scanf("%d %d",&x,&y);
		fa[find(x)]=fa[find(y+O)];
	}
	for(int i=1;i<=O;i++)
		cnt1[find(i)]++;
	for(int i=O+1;i<=O*2;i++)
		cnt2[find(i)]++;
	ll ans=0;
	for(int i=1;i<=O*2;i++)
		ans+=cnt1[i]*cnt2[i];
	printf("%lld\n",ans-N);
	return 0;
}