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 <cstdio>
#include <vector>
#include <queue>
using namespace std;

typedef long long i64;

int main()
{
	const int MAX = 1e5;
	vector<vector<int>> g(2 * MAX);
	int n;

	scanf("%d", &n);
	for(int i = 0; i < n; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		x -= 1;
		y -= 1;
		g[x].push_back(MAX + y);
		g[MAX + y].push_back(x);
	}

	i64 ans = 0;
	vector<bool> visited(2 * MAX);

	for(int s = 0; s < 2 * MAX; s++) {

		if(visited[s]) {
			continue;
		}

		queue<int> q;
		int edges = 0;
		int x_vertices = 0, y_vertices = 0;

		q.push(s);

		while(!q.empty()) {

			int v = q.front();
			q.pop();

			if(visited[v]) {
				continue;
			}
			visited[v] = true;
			if(v < MAX) {
				x_vertices += 1;
			} else {
				y_vertices += 1;
			}

			for(auto& w : g[v]) {
				if(!visited[w]) {
					edges += 1;
				}
				q.push(w);
			}
		}

		ans += x_vertices * (i64)y_vertices - edges;
	}

	printf("%lld\n", ans);

	return 0;
}
