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;
long long n,now,X[100005],Y[100005],H[100005],q,ans;
struct V {
	long long x,y;
} A[100005];
struct X {
	long long an1,an2,f;
} Q[200005];
long long Find(long long i) {
	if(Q[i].f==i)return Q[i].f;
	return Q[i].f=Find(Q[i].f);
}
void BU(long long a,long long b) {
	a=Find(a),b=Find(b);
	if(a^b) {
		Q[a].an1+=Q[b].an1;
		Q[a].an2+=Q[b].an2;
		Q[b].f=a;
	}
}
int main() {
	scanf("%lld",&n);
	for(long long i=1; i<=n; i++) {
		scanf("%lld %lld",&A[i].x,&A[i].y);
		if(!X[A[i].x]) {
			X[A[i].x]=++q;
			Q[q].an2=1;
			Q[q].f=q;
		}
		if(!Y[A[i].y]) {
			Y[A[i].y]=++q;
			Q[q].an1=1;
			Q[q].f=q;
		}
	}
	for(long long i=1; i<=n; i++)
		BU(X[A[i].x],Y[A[i].y]);
	for(long long i=1; i<=n; i++){
		q=Find(X[A[i].x]);
		if(!H[q]){
			H[q]=1;
			ans+=Q[q].an1*Q[q].an2;
		}
		q=Find(Y[A[i].y]);
		if(!H[q]){
			H[q]=1;
			ans+=Q[q].an1*Q[q].an2;
		}
	}
	ans-=n;
	printf("%lld",ans);
	return 0;
}