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;
#define ot 100002
#define ll long long
ll n,a,b,c[ot*2],x,y,s;
vector <int> j[ot*2];
void dfs(int p)
{
	c[p]=1,p>ot?y++:x++;
	for(int i=0;i<j[p].size();i++)
		if(!c[j[p][i]])
			dfs(j[p][i]);
}
int main()
{
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>a>>b,j[a].push_back(b+ot),j[b+ot].push_back(a);
	for(int i=0;i<ot*2;i++)
		if(!c[i]&&j[i].size())
			x=y=0,dfs(i),s+=x*y;
	cout<<s-n;
    return 0;
}