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<iostream>
#include<vector>
using namespace std;
#include<vector>
struct UF{
	int n;
	vector<int>parent,rank;
	UF(int n_=0):n(n_),parent(n_),rank(n_,1)
	{
		for(int i=0;i<n_;i++)parent[i]=i;
	}
	int find(int a)
	{
		return parent[a]!=a?parent[a]=find(parent[a]):a;
	}
	bool same(int a,int b)
	{
		return find(a)==find(b);
	}
	bool unite(int a,int b)
	{
		a=find(a),b=find(b);
		if(a==b)return false;
		if(rank[a]<rank[b])
		{
			parent[a]=b;
			rank[b]+=rank[a];
		}
		else
		{
			parent[b]=a;
			rank[a]+=rank[b];
		}
		return true;
	}
};
int N;
vector<int>G[1<<17];
int main()
{
	cin>>N;
	for(int i=0;i<N;i++)
	{
		int x,y;cin>>x>>y;
		x--,y--;
		G[x].push_back(y);
	}
	UF uf(1<<17);
	for(int i=0;i<1<<17;i++)
	{
		for(int j=0;j+1<G[i].size();j++)
		{
			uf.unite(G[i][j],G[i][j+1]);
		}
	}
	long ans=0;
	for(int i=0;i<1<<17;i++)
	{
		if(!G[i].empty())
		{
			ans+=uf.rank[uf.find(G[i][0])];
		}
	}
	cout<<ans-N<<endl;
}
