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;
const int Maxn=100005,maxi=100000;
int fa[2*Maxn],cnt[2*Maxn];
int get_fa(int x)
{
	return x==fa[x]?fa[x]:fa[x]=get_fa(fa[x]);
}
void merge(int x,int y)
{
	int a=get_fa(x),b=get_fa(y);
	if(a!=b)
	{
		cnt[b]+=cnt[a];
		fa[a]=b;
	}
}
int n;
long long tot,ans[2*Maxn][3];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=2*maxi;i++)
		fa[i]=i;
	for(int i=1;i<=n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		merge(x,y+maxi);
		cnt[get_fa(x)]++;
	}
	for(int i=1;i<=maxi;i++)
		ans[get_fa(i)][1]++;
	for(int i=maxi+1;i<=2*maxi;i++)
		ans[get_fa(i)][2]++;
	for(int i=1;i<=2*maxi;i++)
		if(get_fa(i)==i) tot+=(ans[i][1]*ans[i][2]-cnt[i]);
	printf("%lld",tot);
	return 0;
}