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;

const int V = 100005;
vector<int> to[V * 2];
bool visited[V * 2];

vector<int> cnt;
void dfs(int v)
{
	//cout << "dfs(" << v << ")" << endl;
	if (visited[v])
	{
		//cout << v << " is visited(in dfs)." << endl;
		return;
	}
	visited[v] = true;
	cnt[v / V]++;
	for (int u : to[v])
		dfs(u);
}

int main(void)
{
	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])
		{
			//cout << i << " is visited at first." << endl;
			//cout << i << ":" << cnt[0] << " " << cnt[1] << endl;
			continue;
		}
		cnt = vector<int>(2);
		dfs(i);
		ans += (long long)cnt[0] * cnt[1];
		//cout << i << ":" << cnt[0] << " " << cnt[1] << endl;
	}
	ans -= n;
	cout << ans << endl;

	return 0;
}
