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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ABC131
{
    class F
    {
        static void Main(string[] args)
        {
            int N = int.Parse(Console.ReadLine());
            UnionFind uf = new UnionFind(N);

            HashSet<int> xs = new HashSet<int>();
            HashSet<int> ys = new HashSet<int>();
            HashSet<int> xs2 = new HashSet<int>();
            HashSet<int> ys2 = new HashSet<int>();
            for (int i = 0; i < N; i++)
            {
                int[] a = Console.ReadLine().Split(' ').Select(v => int.Parse(v)).ToArray();
                int x = a[0], y = a[1];
                if (xs.Contains(x))
                {
                    xs2.Add(x);
                }
                if (ys.Contains(y))
                {
                    ys2.Add(y);
                }
                xs.Add(x);
                ys.Add(y);
            }

            long r = xs.Count() * (long)ys.Count() - N;
            if (xs.Count() == 0 || ys.Count() == 0) r = 0;
            Console.WriteLine(r);
        }

    }

    class UnionFind
    {
        private int N;
        private int[] parents; //親の番号を格納. 親の場合は-(メンバサイズ)

        public UnionFind(int n)
        {
            parents = new int[n];
            for (int i = 0; i < n; i++) parents[i] = -1;
            N = n;
        }

        //グループのボスを取得
        public int Root(int x)
        {
            if (parents[x] < 0) return x;
            parents[x] = Root(parents[x]); //次からはボスに直結するように構造を変える
            return parents[x];
        }

        //グループのメンバ数
        public int Size(int x)
        {
            int r = Root(x);
            return -parents[r];
        }

        public bool Unite(int x, int y)
        {
            x = Root(x);
            y = Root(y);
            if (x == y) return false;
            //メンバサイズが大きい方をボスとして合併
            if (Size(x) < Size(y))
            {
                parents[y] += parents[x];
                parents[x] = y;
            }
            else
            {
                parents[x] += parents[y];
                parents[y] = x;
            }
            return true;
        }
    }

}