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 ll long long
const int N=1e5+10;
int n,x[N],y[N],fa[N],sz[N];
map<int,int>X,Y,cx[N],cy[N];
int find(int x)
{
	if(fa[x]==x)return x;
	fa[x]=find(fa[x]);
	return fa[x];
}
int main()
{
    scanf("%d",&n);
    int fx,fy;
    for(int i=1;i<=n;i++)
	{
        fa[i]=i;
        sz[i]=1;
        scanf("%d%d",&x[i],&y[i]);
        if(!X.count(x[i]))X[x[i]]=i;
        fx=find(X[x[i]]);
        if(!Y.count(y[i]))Y[y[i]]=i;
        fy=find(Y[y[i]]);
        if(fx!=fy)
		{
            if(sz[fx]<sz[fy])swap(fx,fy);
            fa[fy]=fx;
            sz[fx]+=sz[fy];
            X[x[i]]=fx;
            Y[y[i]]=fx;
        }
    }
    ll ans=0;
    for(int i=1;i<=n;i++)
	{
        fx=find(X[x[i]]);
        cx[fx][x[i]]++;
        cy[fx][y[i]]++;
    }
    for(int i=1;i<=n;i++)
        if(fa[i]==i)
            ans=ans+1ll*cx[i].size()*cy[i].size()-sz[i];
    cout<<ans;
    return 0;
}