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 pb push_back
#define ll long long
#define mp make_pair
#define inf 1000000007
#define int long long
using namespace std;
vector<int> p[214514];
int iro[214514];
signed main(){
	int n;
	cin>>n;
	for(int i=0;i<n;i++){
		int x,y;
		cin>>x>>y;
		y+=100000;
		p[x-1].pb(y-1);
		p[y-1].pb(x-1);
	}
	ll ans = -n;
	for(int i=0;i<200000;i++){
		if(iro[i]==0){
			iro[i]=1;
			int cx=0,cy=0;
			queue<int> que;
			que.push(i);
			if(i<100000)cx++;else cy++;
			while(!que.empty()){
				int ci = que.front();que.pop();
				for(int j=0;j<p[ci].size();j++){
					int cd = p[ci][j];
					if(iro[cd]==0){
						iro[cd]=1;
						if(cd<100000)cx++;else cy++;
						que.push(cd);
					}
				}
			}
			ans += cx*cy;
		}
	}
	cout<<ans<<endl;
	return 0;
}