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 <algorithm>
#include <iostream>
#include <vector>
using namespace std;

const int MAXN = 1e5 + 10, N = 1e5;
vector<int> Mat[MAXN], d[MAXN], r[MAXN], com;
bool se[MAXN];
long long en;
int n;
void dfs(int v) {
	com.push_back(v);
	se[v] = 1;
	for (int i = 0; i < Mat[v].size(); i++)
		if (!se[Mat[v][i]])
			dfs(Mat[v][i]);
}

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		int a, b;
		cin >> a >> b;
		d[b].push_back(a);
		r[a].push_back(b);
	}
	for (int i = 1; i <= N; i++)
		for (int j = 1; j < d[i].size(); j++)
			Mat[d[i][j]].push_back(d[i][0]), Mat[d[i][0]].push_back(d[i][j]);
	for (int i = 1; i <= N; i++) {
		if (se[i])
			continue;
		com.clear();
		dfs(i);
		vector<int> x;
		for (int j = 0; j < com.size(); j++)
			for (int k = 0; k < r[com[j]].size(); k++)
				x.push_back(r[com[j]][k]);
		sort(x.begin(), x.end());
		x.resize(unique(x.begin(), x.end()) - x.begin());
		en += (long long)x.size() * (long long)com.size();
	}
	cout << en - n;
}