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<iostream>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=100005;

struct point{
	int x,y,id;
}p[maxn];

int f[maxn];

bool cmpx(point a, point b){return a.x<b.x;}
bool cmpy(point a, point b){return a.y<b.y;}
int G(int x){ return f[x]==x?x:f[x]=G(f[x]); }
void U(int x, int y){
	int fx=G(x), fy=G(y);
	if (fx!=fy){ f[fx]=fy;}
}
set<int> SX[maxn],SY[maxn];
int main(){
	int n;
	scanf("%d",&n);
	for (int i=1;i<=n;i++) f[i]=i;
	for (int i=1;i<=n;i++)scanf("%d%d",&p[i].x,&p[i].y),p[i].id=i;
	sort(p+1,p+1+n,cmpx);
	for (int i=2;i<=n;i++){if (p[i].x==p[i-1].x) U(p[i].id,p[i-1].id); }
	sort(p+1,p+1+n,cmpy);
	for (int i=2;i<=n;i++){if (p[i].y==p[i-1].y) U(p[i].id,p[i-1].id); }
	for (int i=1;i<=n;i++){
		int fi=G(p[i].id);
		SX[fi].insert(p[i].x); SY[fi].insert(p[i].y);
	}
	long long  ans=0;
	for (int i=1;i<=n;i++){
		ans+=(SX[i].size() * SY[i].size());
	}
	printf("%lld",ans-n);
	return 0; 
}