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>
using namespace std;
long long a,b,c,d,e,i,j,vl=100000,msh[200009],pas;
vector <long long> v[200009];
bool bo[200009];
long long fnd(long long q){
	if(msh[q]==q) return q; else return msh[q]=fnd(msh[q]);
}
void mrg(long long q, long long w){
	q=fnd(q);w=fnd(w);
	if(q==w) return;
	if(v[q].size()<v[w].size()) swap(q,w);
	msh[w]=q;
	for(long long h=0; h<v[w].size(); h++){
		v[q].push_back(v[w][h]);
	}
}
int main(){
	cin>>a;
	for(i=1; i<=vl*2; i++){
		v[i].push_back(i);msh[i]=i;
	}
	for(i=1; i<=a; i++){
		cin>>c>>d;
		mrg(c,d+vl);
	}
	for(i=1; i<=vl*2; i++){
		c=fnd(i);
		if(bo[c]==1) continue;
		bo[c]=1;
		d=0;e=0;
		for(j=0; j<v[c].size(); j++){
			if(v[c][j]<=vl) d++; else e++;
		}
		pas+=d*e;
	}
	cout<<pas-a;
	return 0;
}