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<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<iomanip>
#include<set>

using namespace std;

const int V=100005;

vector<int> to[V*2];

vector<int> cnt;

vector<int> visited(V*2);

void dfs(int v)
{
	if(visited[v])return;
	visited[v]=true;
	cnt[v/V]++;
	for(auto u:to[v])
		dfs(u);


}

int main()
{
	int n;
	cin >> n;
	for(int i=0;i<n;i++)
	{
		int x,y;
		cin >> x >> y;
		y+=V;
		to[x].push_back(y);
		to[y].push_back(x);
	}


	long long ans=0;
	for(int i=0;i<V*2;i++)
	{
		if(visited[i])continue;
		cnt=vector<int>(2,0);
		dfs(i);
		ans+=(long long)cnt[0]*cnt[1];
	}

	ans-=n;
	
	cout << ans << endl;



}