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 int
#define pb push_back
using namespace std;
ll a,b,visit[500005];
vector<ll> v[500005];
void dfs(ll cur)
{
	visit[cur]=1;
	if(cur>100000) a++;
	else b++;
	for(auto x:v[cur]) if(!visit[x]) dfs(x);
}
int main()
{
	ll n,i,x,y;
	cin>>n;
	for(i=0;i<n;i++)
	{
		cin>>x>>y;
		v[x].pb(y+100000);
		v[y+100000].pb(x);
	}
	ll s=0;
	for(i=1;i<=200000;i++)
	{
		if(!visit[i])
		{
			a=0;
			b=0;
			dfs(i);
			s+=a*b;
		}
	}
	cout<<s-n;
	return 0;
}