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;
typedef long long ll;
#define int ll
int n;
vector<bool> vis;
vector<vector<int>> g;
const int N=1e5;
int f=0,h=0;
void dfs(int node)
{
	vis[node]=1;
	if(node<N)f++;
	else h++;
	for(auto u:g[node])
	{
		if(!vis[u])dfs(u);
	}
}
int32_t main()
{
	#ifndef ONLINE_JUDGE
	freopen("in","r",stdin);
	freopen("out","w",stdout);
	#endif
	cin>>n;
	int u,v;
	g=vector<vector<int>>(2*N+1);
	vis=vector<bool>(2*N+1);
	for(int i=0;i<n;i++)
	{
		cin>>u>>v;
		u--;v--;
		v+=N;
		g[u].push_back(v);
		g[v].push_back(u);
	}

	
	int ans=0;

	for(int i=0;i<N;i++)
	{
		if(!vis[i]){f=0,h=0;dfs(i);ans+=f*h;}
	}
	cout<<ans-n;




	

}