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

vector<int> adj[200001];
int vis[200001];
ll cntx,cnty;

void dfs(int v){
	vis[v]=1;
	if(v>100000) cnty++;
	else cntx++;
	for(int u:adj[v]){
		if(vis[u]) continue;
		dfs(u);
	}
	return;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);

	int n,x,y;
	ll ans=0;
	cin>>n;
	memset(vis,0,sizeof(vis));
	for(int i=0;i<n;i++){
		cin>>x>>y;
		adj[x].push_back(100000+y);
		adj[100000+y].push_back(x);
	}
	for(int i=1;i<=100000;i++){
		if(adj[i].size()){
			if(vis[i]==0){
				cntx=cnty=0;
				dfs(i);
				ans+=(cntx*cnty);
			}
		}
	}
	cout<<ans-n<<"\n";
	return 0;
}