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.*;

public class Main {

    int n;
    int[] xs, ys;

    int cnt;
    E[] g;

    void run() {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        xs = new int[n];
        ys = new int[n];

        TreeSet<Integer> xp = new TreeSet<>(), yp = new TreeSet<>();

        for (int i = 0; i < n; i++) {
            xs[i] = sc.nextInt();
            ys[i] = sc.nextInt();
            xp.add(xs[i]);
            yp.add(ys[i]);
        }

        HashMap<Integer, Integer> x2int = new HashMap<>(), y2int = new HashMap<>();

        cnt = 0;
        for (int p: xp) x2int.put(p, cnt++);
        for (int p: yp) y2int.put(p, cnt++);

        g = new E[cnt];
        for (int i = 0; i < cnt; i++) g[i] = new E();

        for (int i = 0; i < n; i++) {
            int xidx = x2int.get(xs[i]), yidx = y2int.get(ys[i]);
            g[xidx].add(yidx);
            g[yidx].add(xidx);
        }

        long ans = 0;
        boolean[] visited = new boolean[cnt];
        for (int i = 0; i < cnt; i++) if (!visited[i]) {
            Ret r = solve(i, true, visited);
            long max = Math.max(r.xc, r.yc);
            ans += max * max - r.ec;
        }
        System.out.println(ans);
    }


    Ret solve(int cur, boolean isX, boolean[] visited) {
        visited[cur] = true;
        Ret ret = new Ret();

        for (int next: g[cur]) if (!visited[next]) {
            Ret n = solve(next, !isX, visited);
            ret.ec += n.ec + 1;
            ret.xc += n.xc;
            ret.yc += n.yc;
        }
        if (isX) ret.xc++;
        else ret.yc++;
        return ret;
    }

    class Ret {
        int ec, xc, yc;
        Ret() {
            this.ec = 0;
            this.xc = 0;
            this.yc = 0;
        }
    }

    class E extends ArrayList<Integer> {}


    void debug(Object...os) {
        System.err.println(Arrays.deepToString(os));
    }

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