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<bits/stdc++.h>
#define LL long long
#define P 1000000007
#define M 1000005
using namespace std;
LL ans=0;
int mkx[M],mky[M],tx,ty,x[M],y[M];
vector<int>edge[M];
bool mark[M];
int L,R,E;
void dfs(int x) {
	E++;
	if(mark[x])return;
	mark[x]=1;
	if(x<=tx)L++;
	else R++;
	for(int i=0; i<edge[x].size(); i++)
		dfs(edge[x][i]);
}
int main() {
	int n;
	cin>>n;
	for(int i=1; i<=n; i++) {
		scanf("%d %d",&x[i],&y[i]);
		if(!mkx[x[i]])mkx[x[i]]=++tx;
		if(!mky[y[i]])mky[y[i]]=++ty;
	}
	for(int i=1; i<=n; i++) {
		int a=mkx[x[i]];
		int b=mky[y[i]]+tx;
		edge[a].push_back(b);
		edge[b].push_back(a);
	}
	for(int i=1; i<=tx+ty; i++) {
		if(!mark[i]) {
			L=0,R=0,E=0;
			dfs(i);
			if(L<2||R<2)continue;
			E/=2;
			ans+=1LL*L*R-E;
		}
	}
	cout<<ans<<endl;
	return 0;
}