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: Wrong Answer)

import java.util.*;
class Main {
    static int cntX;
    static int cntY;
    static HashMap<Integer, List<Integer>> map;
    static boolean[] visited;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        map = new HashMap<>();

        for(int i = 0; i<N; i++) {
            int x = sc.nextInt();
            int y = sc.nextInt()+100001;
            if(!map.containsKey(x)) {
                map.put(x, new ArrayList<>());
            }
            if(!map.containsKey(y)) {
                map.put(y, new ArrayList<>());
            }
            map.get(x).add(y);
            map.get(y).add(x);
        }

        visited = new boolean[100001*2];
        long ans = 0;
        for(int i = 0; i<100001; i++) {
            cntX = 0;
            cntY = 0;
            dfs(i);
            ans += cntX*cntY;
        }

        System.out.println(ans-N);

    }
    public static void dfs(int x) {
        if(visited[x]) return;
        visited[x] = true;
        if(x<100001) cntX++;
        else cntY++;
        if(map.containsKey(x)) {
            for(int i: map.get(x)) {
                dfs(i);
            }
        }
    }
}