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 N 100005
using namespace std;
int n,f[N],x[N],y[N];
vector<int>X[N];
set<int>S[N];
int find(int x) {
	return f[x]==x?x:f[x]=find(f[x]);
}
int main() {
	scanf("%d",&n);
	for(int i=1; i<=N-5; i++)f[i]=i;
	for(int i=1; i<=n; i++) {
		scanf("%d%d",&x[i],&y[i]);
		X[x[i]].push_back(y[i]);
	}
	for(int i=1; i<=N-5; i++) {
		int fa;
		if(X[i].size())fa=find(X[i][0]);
		for(int j=1; j<X[i].size(); j++) {
			int x=find(X[i][j]);
			f[x]=fa;
		}
	}
	for(int i=1; i<=N-5; i++) {
		int fa;
		if(X[i].size())fa=find(X[i][0]);
		for(int j=0; j<X[i].size(); j++)
			S[fa].insert(X[i][j]);
	}
	long long ans=0;
	for(int i=1; i<=N-5; i++) {
		if(X[i].size()) {
			int fa=find(X[i][0]);
			ans+=1ll*S[fa].size();
		}
	}
	printf("%lld\n",ans-n);
	return 0;
}
