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>
#define maxn 100010
using namespace std;
int n,cnt,f[maxn<<1],sx[maxn<<1],sy[maxn<<1],mp1[maxn],mp2[maxn];
inline int find(int x){return x==f[x]?x:f[x]=find(f[x]);}
inline void link(int x,int y){
	x=find(x),y=find(y);
	if(x==y)return;
	f[x]=y,sx[y]+=sx[x],sy[y]+=sy[x]; 
}
int main(){
	scanf("%d",&n);
	cnt=0;
	for(int i=1;i<=n;++i){
		int x,y;
		scanf("%d%d",&x,&y);
		if(!mp1[x]){sx[mp1[x]=++cnt]=1;f[cnt]=cnt;}
		if(!mp2[y]){sy[mp2[y]=++cnt]=1;f[cnt]=cnt;}
		link(mp1[x],mp2[y]);
	}
	long long ans=-n;
	for(int i=1;i<=cnt;++i)if(f[i]==i)ans+=1ll*sx[i]*sy[i];
	printf("%lld\n",ans);
	return 0;
}