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: Accepted)

using System;
using System.Linq;//リストの使用
using System.Collections.Generic;
class Program
{
static long n = 200000;
static long q = long.Parse(Console.ReadLine());
static long[] parents = new long[n];//親
static long[] edgeCounts = new long[n];
static List<long>[] parentList = new List<long>[n];
  
	static void Main()
	{
    long answer = 0;
    
    for(int i = 0; i < (int)n; i++) parentList[i] = new List<long>();//リストの宣言
    for(long i = 1; i < n; i++)
    {
      parents[i] = i;
    }
    
    for(long i = 0; i < q; i++)
    {
      long[] nums = Array.ConvertAll(Console.ReadLine().Split(' '),long.Parse);
      edgeCounts[nums[0]-1]++;
      if(whatParent(nums[0]-1) != whatParent(nums[1]-1+100000))
        parents[whatParent(nums[0]-1)] = whatParent(nums[1]-1+100000);//親を結合して統一。親が2人できることはない。
    }

    for(int i = 0; i < n; i++) parentList[whatParent(i)].Add(i);

    for(int i = 0; i < n; i++)
    {
      long x = 0;
      long y = 0;
      long ll = 0;
      for(int j = 0; j < parentList[i].Count(); j++)
      {
        if(parentList[i][j] < 100000) x++;
        else y++;

        ll += edgeCounts[parentList[i][j]];
      }
      answer += x*y-ll;
    }
    Console.WriteLine(answer);
	}

  static long whatParent(long child)
  {
    if(parents[child] == child) return child;
    else return parents[child] = whatParent(parents[child]);
  }
  
}