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 mp make_pair
typedef long long ll;
typedef pair<int,int>pii;
const int N = 1e5;
vector<int>graph[2*N+100];
vector<int>comp1,comp2;
int vis[2*N+100];
void dfs(int u)
{
	vis[u] = 1;
	if(u<=N){
		comp1.push_back(u);
	}
	else comp2.push_back(u);
	for(auto it:graph[u])
	{
		if(!vis[it]){
			dfs(it);
		}
	}
}
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		graph[x].push_back(y+N);
		graph[y+N].push_back(x);
	}
	ll ans = 0;
	for(int i=1;i<=2*N;i++){
		if(!vis[i]){
			comp1.clear();
			comp2.clear();
			dfs(i);
			ll cnt = 0;
			for(auto it:comp1)
			{
				cnt += graph[it].size();
			}
			for(auto it:comp2)
			{
				cnt += graph[it].size();
			}
			cnt /= 2;
			ans += (1ll*comp1.size()*comp2.size()-cnt);
		}
	}
	printf("%lld\n",ans);
}
