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 M 100000
#define N 200005
using namespace std;
vector<int>son[N];
int E,L,R,mark[N];
void dfs(int x,int flg){
	E++;
	if(mark[x])return;
	mark[x]=1;
	if(flg)L++;
	else R++;
	for(int i=0;i<son[x].size();i++){
		int y=son[x][i];
		dfs(y,!flg);
	}
}
int main(){
	int n,x,y;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d%d",&x,&y),son[x].push_back(y+M),son[y+M].push_back(x);
	long long ans=0;
	for(int i=1;i<=M;i++){
		if(!mark[i]){
			E=L=R=0;
			if(son[i].size()==0)continue;
			dfs(i,1);
			if(L<2||R<2)continue;
			E/=2;
			ans+=1ll*L*R-E;
		}
	}
	printf("%lld\n",ans);
	return 0;
}