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>
#define For(a,b,c) for(int a=b;a<=c;++a)
#define Dor(a,b,c) for(int a=b;a>=c;--a)
using namespace std;
const int N=300300;
int n,C[N],S[N],X[N],Y[N];
int Find(int u) {
	return u==C[u]?u:C[u]=Find(C[u]);
}
void Uni(int x,int y) {
	x=Find(x),y=Find(y);
	if(x!=y) C[x]=y,S[y]+=S[x],X[y]+=X[x],Y[y]+=Y[x];
}
int main() {
	scanf("%d",&n);
	For(i,1,300020) C[i]=i;
	For(i,1,100000) X[i+100010]=1;
	For(i,1,100000) Y[i+200020]=1;
	For(i,1,n) S[i]=1;
	For(i,1,n) {
		int x,y;
		scanf("%d%d",&x,&y);
		Uni(i,x+100010);
		Uni(i,y+200020);
	}
	long long ans=0;
	For(i,1,300020)
	if(i==Find(i)) {
		int x=X[i],y=Y[i],s=S[i];
		ans+=1LL*x*y-s;
	}
	printf("%lld",ans);
	return 0;
}