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
#define pb push_back

const int maxn = 1e5 + 20;

vector<int> adj[2 * maxn];

int ta , tb;

bool visited[maxn * 2];

void dfs(int v)
{
	if(v < maxn)
		ta++;
	else
		tb++;
	visited[v] = 1;

	for(auto u : adj[v])
		if(!visited[u])
			dfs(u);
}

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

	int n;
	cin >> n;

	for(int i = 0; i < n; i++)
	{
		int x , y;
		cin >> x >> y;
		y += maxn;

		adj[x].pb(y);
		adj[y].pb(x);
	}

	ll res = -n;
	for(int i = 0; i < 2 * maxn; i++)
		if(!visited[i])
		{
			ta = tb = 0;
			dfs(i);
			res += 1LL * ta * tb;
		}

	cout << res << endl;
}










 
