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 int long long
#define pb push_back
using namespace std;

const int N=1e5+5;
int n, vis[N], ans, cnt, x, y;
vector<int> pts[N], ony[N];
vector<int> adj[N];
set<int> ys;
void dfs(int u)
{
	vis[u]=1;
	cnt++;
	for(auto p: pts[u])
	{
		ys.insert(p);
	}
	for(auto v: adj[u])
	{
		if(!vis[v])
		{
			dfs(v);
		}
	}
}
signed main()
{
	cin >> n;
	for(int i = 0; i < n; i++)
	{
		cin >> x >> y;
		ony[y].pb(x);
		pts[x].pb(y);
	}
	for(int i = 0; i < N; i++)
	{
		for(int j = 0; j < (int) ony[i].size()-1; j++)
		{
			adj[ony[i][j]].pb(ony[i][j+1]);
			adj[ony[i][j+1]].pb(ony[i][j]);
		}
	}
	for(int i = 0; i < N; i++)
	{
		if(!vis[i]) dfs(i);
		ans+=ys.size()*cnt;
		ys.clear();
		cnt=0;
	}
	cout << ans-n;
}