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<algorithm>
#include<cstdio>
#include<vector>
#include<set>
#define mxn 100010
#define LL long long
using namespace std;
const int M=1e5;
int n,sl,fh,fa[mxn];
LL ans;
set<int> st[mxn];
vector<int> vt[mxn];
int rd()
{
	sl=0;fh=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch) {if(ch=='-') fh=-1; ch=getchar();}
	while('0'<=ch&&ch<='9') sl=sl*10+ch-'0',ch=getchar();
	return sl*fh;
}
int getfa(int x)
{
	if(fa[x]!=x) fa[x]=getfa(fa[x]);
	return fa[x];
}
int main()
{
	n=rd();int x,y,fx;
	for(int i=1;i<=n;++i) x=rd(),y=rd(),vt[x].push_back(y);
	for(int i=1;i<=M;++i) fa[i]=i;
	for(int sz,i=1;i<=M;++i)
	{
		sz=vt[i].size();if(sz) fx=getfa(vt[i][0]);
		for(int j=1;j<sz;++j) fa[getfa(vt[i][j])]=fx;
	}
	for(int sz,i=1;i<=M;++i)
	{
		sz=vt[i].size();
		for(int j=0;j<sz;++j) st[getfa(vt[i][j])].insert(i);
	}
	for(int i=1;i<=M;++i) ans+=st[getfa(i)].size();
	printf("%lld\n",ans-n);
	return 0;
}