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.Linq;
using System.Collections.Generic;
using static System.Console;
using System.Text;

namespace AtCoder_Beginner_Contest_131
{
    //参照(http://drken1215.hatenablog.com/entry/2019/06/23/004700)
    class Program
    {
        class UnionFind
        {
            int[] par;
            public UnionFind(int n)
            {
                init(n);
            }
            public void init(int n)
            {
                par = new int[n];
                for (var i = 0; i < n; ++i)
                { par[i] = -1; }
            }
            public int root(int x)
            {
                if (par[x] < 0) return x;
                else return par[x] = root(par[x]);
            }
            public bool issame(int x, int y)
            {
                return root(x) == root(y);
            }
            public bool merge(int x, int y)
            {
                x = root(x); y = root(y);
                if (x == y) return false;
                if (par[x] > par[y])
                {
                    var tmp = y;
                    y = x;
                    x = tmp;
                }
                par[x] += par[y];
                par[y] = x;
                return true;
            }
            public int size(int x)
            {
                return -par[root(x)];
            }
        }
        const int MAX = 110000;
        static public long[] Sarray() { return ReadLine().Split().Select(long.Parse).ToArray(); }
        static public List<long> Slist() { return ReadLine().Split().Select(long.Parse).ToList(); }
        static void Main(string[] args)
        {

            var N = Sarray()[0];
            UnionFind uf = new UnionFind(MAX * 2);
            var x = new List<int>();
            var y = new List<int>();
            for (var i = 0; i < N; ++i)
            {
                var xy = Sarray();
                x.Add((int)xy[0]);
                y.Add((int)xy[1]);
                uf.merge(x[i], y[i] + MAX);
            }
            var mx = new int[MAX * 2];
            var my = new int[MAX * 2];
            for (var i = 0; i < MAX; ++i) { mx[uf.root(i)]++; }
            for (var i = MAX; i < MAX * 2; ++i) { my[uf.root(i)]++; }
            int res = 0;
            for (var r = 0; r < MAX * 2; ++r) res += mx[r] * my[r];
            WriteLine(res - N);
        }
    }
}