Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 1
5 1
5 5

Sample Output 1

1

By choosing a = 1, b = 1, c = 5, d = 5, we can add a dot at (1, 5). We cannot do the operation any more, so the maximum number of operations is 1.

Sample Input 2

2
10 10
20 20

Sample Output 2

0

There are only two dots, so we cannot do the operation at all.

Sample Input 3

9
1 1
2 1
3 1
4 1
5 1
1 2
1 3
1 4
1 5

Sample Output 3

16

We can do the operation for all choices of the form a = 1, b = 1, c = i, d = j (2 \leq i,j \leq 5), and no more. Thus, the maximum number of operations is 16.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.util.*;

class Main {

	private static final int V = 100005;

	private static List<Integer>[] G;
	private static boolean[] visited;
	private static long[] count = new long[2];

	protected static void dfs(int v) {
		if (visited[v]) return;
		visited[v] = true;
		count[v / V]++;
		for (int nv : G[v]) dfs(nv);
	}

	public static void main(String[] args) throws Exception {
		final Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();

		G = new ArrayList[V * 2];
		for (int i = 0; i < V * 2;
			 i++) {
			G[i] = new ArrayList<>();
		}

		for (int i = 0; i < N; i++) {
			int x = sc.nextInt();
			int y = sc.nextInt() + V;
			G[x].add(y);
			G[y].add(x);
		}

		long answer = 0;
		visited = new boolean[V * 2];
		for (int i = 0; i < V * 2; i++) {
			if (visited[i]) continue;
			dfs(i);
			answer += count[0] * count[1];
			count[0] = count[1] = 0;
		}

		System.out.println(answer - N);
	}
}