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
import java.util.*;

class Main {
	static int V = 100005;
	static boolean[] visited = new boolean[V*2];
	static ArrayList<LinkedList<Integer>> _to;
	static int[] cnt;
	static void dfs(int v) {
		if (visited[v]) {
			return;
		}
		visited[v]=true;
		cnt[v/V]++;
		for (Integer u: _to.get(v)) {
			dfs(u);
		}
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = Integer.parseInt(sc.next());
		_to = new ArrayList<LinkedList<Integer>>();
		for (int i=0;i<V*2;i++) {
			LinkedList<Integer> tmp = new LinkedList<Integer>();
			_to.add(tmp);
		}
		for (int i=0;i<N;i++) {
			int x=Integer.parseInt(sc.next());
			int y=Integer.parseInt(sc.next());
			y+=V;
			_to.get(x).add(y);
			_to.get(y).add(x);
		}
		long ans = 0L;
		for (int i=0;i<2*V;i++) {
			if (visited[i]) {
				continue;
			}
			cnt = new int[2];
			dfs(i);
			ans += (long)cnt[0]*cnt[1];
		}
		ans-=N;
		System.out.println(ans);
	}
}