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;
#define ll long long
const int N=100005;
const ll p=998244353ll;
template<class IT>inline void cmin(IT &a,IT b){if(b<a)a=b;}
template<class IT>inline void cmax(IT &a,IT b){if(a<b)a=b;}
int g[N],h[N],f[N],sz[N];
int find(int v){return f[v]?f[v]=find(f[v]):v;}
int a[N],b[N];
int main(){
	int n,i,j,x,y,v,u;ll s=0ll;
	scanf("%d",&n);
	for(i=1;i<=n;++i)sz[i]=1;
	for(i=1;i<=n;++i){
		scanf("%d%d",&x,&y);
		if(a[x]){
			v=find(a[x]);
			u=find(i);
			if(v!=u){f[u]=v;sz[v]+=sz[u];}
		}
		a[x]=i;
		if(b[y]){
			v=find(b[y]);
			u=find(i);
			if(v!=u){f[u]=v;sz[v]+=sz[u];}
		}
		b[y]=i;
	}
	for(i=1;i< N;++i)if(a[i])++g[find(a[i])];
	for(i=1;i< N;++i)if(b[i])++h[find(b[i])];
	for(i=1;i<=n;++i)s+=((ll)g[i])*h[i];
	printf("%lld",s-n);
	return 0;
}