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.util.*;

public class Main {

    int N = 2 * 100000;

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

        UnionFind uf = new UnionFind(N);

        for (int i = 0; i < n; i++) {
            xs[i] = sc.nextInt() - 1;
            ys[i] = sc.nextInt() - 1 + N / 2;
            uf.union(xs[i], ys[i]);
        }

        HashMap<Integer, Integer> xc = new HashMap<>();
        HashMap<Integer, Integer> yc = new HashMap<>();

        for (int i = 0; i < N / 2; i++) {
            xc.put(uf.find(i), xc.getOrDefault(uf.find(i), 0) + 1);
            yc.put(uf.find(i + N / 2), yc.getOrDefault(uf.find(i + N / 2), 0) + 1);
        }

        long ans = 0;
        for (int i = 0; i < N; i++) {
            ans += 1L * xc.getOrDefault(i, 0) * yc.getOrDefault(i, 0);
        }
        System.out.println(ans - n);
    }

    class UnionFind {
        private int n, t;
        private int[] parent, rank;

        public UnionFind(int n) {
            this.n = n;
            this.parent = new int[n];
            this.rank = new int[n];
            Arrays.fill(rank, 1);
            for (int i = 0; i < n; i++) parent[i] = i;
        }

        public void union(int x, int y) {
            t++;
            int xr = find(x);
            int yr = find(y);

            if (xr > yr) {
                parent[yr] = xr;
            } else if (xr < yr) {
                parent[xr] = yr;
            } else {
                parent[yr] = xr;
                rank[yr]++;
            }

        }

        public int find(int x) {
            if (x != parent[x]) {
                parent[x] = find(parent[x]);
            }
            return parent[x];
        }

        public boolean same(int x, int y) {
            return find(x) == find(y);
        }

    }

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

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