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;
const int maxn=1e5;
const int N=maxn+10;
int n,par[2*N];
int cntx[2*N],cnty[2*N],sz[2*N];
int x,y;
ll ans;
int find(int x)
{
	return par[x]==x?x:par[x]=find(par[x]);
}
void merge(int x,int y)
{
	x=find(x),y=find(y);
	if(x==y)return;
	if(sz[x]<sz[y])swap(x,y);
	sz[x]+=sz[y];
	cntx[x]+=cntx[y];
	cnty[x]+=cnty[y];
	par[y]=x; 
}
int main()
{
	for(int i=1;i<=maxn;++i)
	par[i]=i,sz[i]=1,cntx[i]=1;
	for(int i=maxn+1;i<=2*maxn;++i)
	par[i]=i,sz[i]=1,cnty[i]=1;
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
	{
		scanf("%d%d",&x,&y);
		y+=maxn;
		merge(x,y);
	}
	for(int i=1;i<=2*maxn;++i)
	{
		x=find(i);
		if(x!=i)continue;
		ans+=1ll*cntx[i]*cnty[i];		
	}
	printf("%lld\n",ans-n);
	return 0;
}