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 M 100005
vector<int>d[2][M];
bool mk1[M],mk2[M],vs[2][M];
int l,r,zs,mx;
void dfs(int x,int p,int dp){
	if(vs[p][x])return;
	vs[p][x]=1;mx=max(mx,dp);
	if(!p)++l;
	else ++r;
	for(int i=0;i<(int)d[p][x].size();++i){
		int v=d[p][x][i];zs++;
		dfs(v,1-p,dp+1);
	}
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1,a,b;i<=n;++i){
		scanf("%d%d",&a,&b);
		d[0][a].push_back(b);
		d[1][b].push_back(a);
		mk1[a]=mk2[b]=1;
	}long long ans=0;
	for(int i=1;i<M;++i)if(mk1[i]){
		l=r=zs=mx=0;dfs(i,0,0);
		ans+=1ll*r*l-zs/2;
	}
	printf("%lld",ans);
}