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;
const int N=2e5+10;
typedef long long LL;
int n,X,Y;
vector<int> G[N],v;
bool vis[N];

void dfs(int x,int dep) {
	if (dep%2) X++; else Y++;
	v.push_back(x);
	vis[x]=1;
	for (int i=0;i<G[x].size();i++) {
		int t=G[x][i];
		if (vis[t]) continue;
		dfs(t,dep+1);
	}
}

int main()
{
	cin>>n;
	for (int i=1;i<=n;i++) {
		int x,y; scanf("%d%d",&x,&y);
		G[x].push_back(y+100000);
		G[y+100000].push_back(x);
	}
	
	memset(vis,0,sizeof(vis));
	LL ans=0;
	for (int i=1;i<=200000;i++) 
		if (G[i].size() && !vis[i]) {
			X=0; Y=0; v.clear();
			dfs(i,1);
			LL num=0;
			for (int j=0;j<v.size();j++) num+=G[v[j]].size();
			ans+=(LL)X*Y-num/2;
		}
	cout<<ans<<endl;	
	return 0;
}