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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;

public class Main {
	static List<List<Integer>> list;
	static boolean[] visit;
	static int mx = 100001;
	static int mx2 = 200003;

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] sa = br.readLine().split(" ");
		int n = Integer.parseInt(sa[0]);
		list = new ArrayList<>(mx2);
		for (int i = 0; i < mx2; i++) {
			list.add(new ArrayList<>());
		}
		for (int i = 0; i < n; i++) {
			sa = br.readLine().split(" ");
			int x = Integer.parseInt(sa[0]);
			int y = Integer.parseInt(sa[1]) + mx;
			list.get(x).add(y);
			list.get(y).add(x);
		}
		br.close();

		long ans = 0;
		visit = new boolean[mx2];
		for (int i = 1; i <= mx; i++) {
			if (!visit[i]) {
				int[] cnt = dfs(i, 0, new int[2]);
				ans += (long) cnt[0] * cnt[1];
			}
		}
		System.out.println(ans - n);
	}

	static int[] dfs(int cur, int p, int[] cnt) {
		visit[cur] = true;
		cnt[cur / mx]++;
		for (int next : list.get(cur)) {
			if (next != p && !visit[next]) {
				dfs(next, cur, cnt);
			}
		}
		return cnt;
	}
}
