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.Collections.Generic;
using System.IO;
using System.Linq;

namespace AtCoder
{
    class Program
    {
        private int N;
        private int[] X, Y;
        static void Main(string[] args)
        {
            new Program().Solve();
        }

        public void Solve()
        {
            var sc = new Scanner();
            N = sc.NextInt();
            X = new int[N];
            Y = new int[N];

            for (int i = 0; i < N; i++)
            {
                X[i] = sc.NextInt();
                Y[i] = sc.NextInt();
            }
            var uf = new UnionFind(200001);
            for (int i = 0; i < N; i++)
            {
                uf.Union(X[i], Y[i] + 100000);
            }

            // Xの同じ連結成分の個数
            var xhs = new HashSet<int>();
            var yhs = new HashSet<int>();
            var xmap = new HashMap<int, int>();
            var ymap = new HashMap<int,int>();
            for (int i = 0; i < N; i++)
            {
                if(xhs.Add(X[i]))
                    xmap[uf.Find(X[i])]++;

                if(yhs.Add(Y[i]))
                    ymap[uf.Find(Y[i] + 100000)]++;
            }

            long sum = 0;

            foreach (var i in xmap)
            {
                sum += (long) i.Value * ymap[i.Key];
            }
            Console.WriteLine(sum-N);
        }
    }
}

class HashMap<K, V> : Dictionary<K, V>
{
    new public V this[K i]
    {
        get
        {
            V v;
            return TryGetValue(i, out v) ? v : base[i] = default(V);
        }
        set { base[i] = value; }
    }
}

class UnionFind
{
    private int[] Par, Rank;
    public UnionFind(int max)
    {
        Par = new int[max];
        Rank = new int[max];
        for (int i = 0; i < max; i++)
        {
            Par[i] = i;
        }
    }

    public int Find(int n)
    {
        if (Par[n] == n)
        {
            return n;
        }
        else
        {
            return Find(Par[n]);
        }
    }

    public bool Same(int a, int b)
    {
        return Find(a) == Find(b);
    }

    public void Union(int a, int b)
    {
        a = Find(a);
        b = Find(b);
        if (a == b) return;
        if (Rank[a] < Rank[b])
        {
            Par[a] = b;
        }
        else
        {
            Par[b] = a;
            if (Rank[a] == Rank[b]) Rank[a]++;
        }
    }
}


class Scanner
{
    public Scanner()
    {
        _stream = new StreamReader(Console.OpenStandardInput());
        _pos = 0;
        _line = new string[0];
    }

    const char Separator = ' ';
    private readonly StreamReader _stream;
    private int _pos;
    private string[] _line;

    #region スペース区切りで取得

    public string Next()
    {
        if (_pos >= _line.Length)
        {
            _line = _stream.ReadLine().Split(Separator);
            _pos = 0;
        }

        return _line[_pos++];
    }

    public int NextInt()
    {
        return int.Parse(Next());
    }

    public long NextLong()
    {
        return long.Parse(Next());
    }

    public double NextDouble()
    {
        return double.Parse(Next());
    }

    #endregion

    #region 型変換

    private int[] ToIntArray(string[] array)
    {
        var result = new int[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = int.Parse(array[i]);
        }

        return result;
    }

    private long[] ToLongArray(string[] array)
    {
        var result = new long[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = long.Parse(array[i]);
        }

        return result;
    }

    private double[] ToDoubleArray(string[] array)
    {
        var result = new double[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = double.Parse(array[i]);
        }

        return result;
    }

    #endregion

    #region 配列取得

    public string[] Array()
    {
        if (_pos >= _line.Length)
            _line = _stream.ReadLine().Split(Separator);

        _pos = _line.Length;
        return _line;
    }

    public int[] IntArray()
    {
        return ToIntArray(Array());
    }

    public long[] LongArray()
    {
        return ToLongArray(Array());
    }

    public double[] DoubleArray()
    {
        return ToDoubleArray(Array());
    }

    #endregion
}