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.*;

public class Main {
	public static void main(String[] args) {
		Main main = new Main();
		main.run();
	}

	ArrayList<ArrayList<Integer>> g = new ArrayList<>();
	int V = 100005;
	boolean used[];
	int cnt[];
	private void dfs(int n) {
		ArrayDeque<Integer> st = new ArrayDeque<>();
		st.add(n);
		used[n]=true;
		cnt[n/V]++;
		while(!st.isEmpty()) {
			int s = st.pop();
			ArrayList<Integer> l = g.get(s);
			for(int i=0; i<l.size(); i++) {
				int t = l.get(i);
				if(!used[t]) {
					used[t]=true;
					cnt[t/V]++;
					st.add(t);
				}
			}
		}
	}
	public void run() {
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		for(int i=0; i<V*2; i++) g.add(new ArrayList<>());
		used = new boolean[V*2];
		for(int i=0; i<n; i++) {
			int x = sc.nextInt();
			int y = sc.nextInt();
			y += V;
			g.get(x).add(y);
			g.get(y).add(x);
		}

		long ans = 0;
		for(int i=0; i<V*2; i++) {
			if(!used[i]) {
				used[i]=true;
				cnt = new int[2];
				dfs(i);
				ans += (long)cnt[0]*cnt[1];
			}
		}
		ans -= n;
		System.out.println(ans);
		sc.close();
	}


}
