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 <cstdio>
#include <vector>
#include <algorithm>
 
using namespace std;

const int MAXN = 100000,MAX = 200005;
int f[MAX],difx[MAX],dify[MAX];
inline int FindSet(int a)
{
	if(a == f[a]) return a;
	return f[a] = FindSet(f[a]);
}
int main()
{
    int i,n; 
	scanf("%d",&n);
	for(i = 1;i <= MAXN * 2;i++) f[i] = i;
	for(i = 1;i <= n;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		f[FindSet(u)] = FindSet(v + MAXN);
	}
	for(i = 1;i <= MAXN;i++)
		difx[FindSet(i)]++;
	for(i = MAXN + 1;i <= MAXN * 2;i++)
		dify[FindSet(i)]++;
	long long ans = 0;
	for(i = 1;i <= MAXN * 2;i++)
		ans += 1ll * difx[i] * dify[i];
	printf("%lld",ans - n);
    return 0;
}