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.ArrayDeque;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;

public class Main {
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		Obj[] arr = new Obj[n];
		Map<Integer, List<Obj>> mapx = new HashMap<>();
		Map<Integer, List<Obj>> mapy = new HashMap<>();
		for (int i = 0; i < n; i++) {
			Obj o = new Obj();
			o.x = sc.nextInt();
			o.y = sc.nextInt();
			arr[i] = o;

			List<Obj> listx = mapx.get(o.x);
			if (listx == null) {
				listx = new ArrayList<>();
				mapx.put(o.x, listx);
			}
			listx.add(o);

			List<Obj> listy = mapy.get(o.y);
			if (listy == null) {
				listy = new ArrayList<>();
				mapy.put(o.y, listy);
			}
			listy.add(o);
		}
		sc.close();

		long ans = 0;
		for (int i = 0; i < n; i++) {
			if (!arr[i].used) {
				Queue<Obj> que = new ArrayDeque<>();
				que.add(arr[i]);
				arr[i].used = true;
				Set<Integer> setx = new HashSet<>();
				Set<Integer> sety = new HashSet<>();
				setx.add(arr[i].x);
				sety.add(arr[i].y);
				while (!que.isEmpty()) {
					Obj cur = que.poll();
					for (Obj next : mapx.get(cur.x)) {
						if (!next.used) {
							que.add(next);
							next.used = true;
							sety.add(next.y);
						}
					}
					for (Obj next : mapy.get(cur.y)) {
						if (!next.used) {
							que.add(next);
							next.used = true;
							setx.add(next.x);
						}
					}
				}
				ans += (long) setx.size() * sety.size();
			}
		}
		System.out.println(ans - n);
	}

	static class Obj {
		int x, y;
		boolean used;
	}
}
