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.IO;
using SB = System.Text.StringBuilder;
//using System.Threading.Tasks;
//using System.Text.RegularExpressions;
//using System.Globalization;
//using System.Diagnostics;
using static System.Console;
using System.Numerics;
using static System.Math;
using pair = Pair<int, int>;

class Program
{
    static void Main()
    {
        //SetOut(new StreamWriter(OpenStandardOutput()) { AutoFlush = false });
        new Program().solve();
        Out.Flush();
    }
    readonly Scanner cin = new Scanner();
    readonly int[] dd = { 0, 1, 0, -1, 0 }; //→↓←↑
    readonly int mod = 1000000007;
    readonly int dom = 998244353;
    bool chmax<T>(ref T a, T b) where T : IComparable<T> { if (a.CompareTo(b) < 0) { a = b; return true; } return false; }
    bool chmin<T>(ref T a, T b) where T : IComparable<T> { if (b.CompareTo(a) < 0) { a = b; return true; } return false; }

    void solve()
    {
        const int M = 100000;
        int N = cin.nextint;
        var U = new DisjointSet(2 * M);
        for (int i = 0; i < N; i++)
        {
            U.Unite(cin.nextint - 1, cin.nextint - 1 + M);
        }

        var P = new int[2 * M];
        var Q = new int[2 * M];

        for (int i = 0; i < M; i++)
        {
            P[U[i]]++;
            Q[U[i + M]]++;
        }

        long cnt = 0;
        for (int i = 0; i < 2 * M; i++)
        {
            cnt += P[i] * Q[i];
        }

        WriteLine(cnt - N);
    }

}

/// <OriginalAuthor>camypaper</OriginalAuthor>
class DisjointSet
{
    readonly int[] par;
    readonly byte[] rank;
    /// <summary>
    /// 0 から <paramref name="N"/>-1 までの番号がついた <paramref name="N"/> 個の要素からなる集合を作成します．
    /// </summary>
    /// <param name="N">要素数</param>
    /// <remarks>このコンストラクタは O(N) で実行されます．</remarks>
    public DisjointSet(int N)
    {
        par = new int[N];
        rank = new byte[N];
        for (int i = 0; i < N; i++) par[i] = -1;
    }
    /// <summary>
    /// 指定した要素が属する集合の代表値を取得します．
    /// </summary>
    /// <param name="id">調べたい要素の 0-indexed での番号</param>
    /// <returns>指定した要素が属する集合の代表値</returns>
    /// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
    public int this[int id]
    {
        get
        {
            if (par[id] < 0) return id;
            return par[id] = this[par[id]];
        }
    }
    /// <summary>
    /// 指定した 2 つの要素が属する集合同士を 1 つに統合することを試みます．
    /// </summary>
    /// <param name="x">最初の要素の 0-indexed での番号</param>
    /// <param name="y">2 つ目の要素の 0-indexed での番号</param>
    /// <returns>統合に成功したならば true，そうでなければ false．</returns>
    /// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
    public bool Unite(int x, int y)
    {
        x = this[x]; y = this[y];
        if (x == y) return false;
        if (rank[x] < rank[y]) { var tmp = x; x = y; y = tmp; }
        par[x] += par[y];
        par[y] = x;
        if (rank[x] == rank[y])
            rank[x]++;
        return true;
    }
    /// <summary>
    /// 指定した要素が属する集合のサイズを取得します．
    /// </summary>
    /// <param name="x">指定する要素の 0-indexed での番号</param>
    /// <returns>集合のサイズ</returns>
    /// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
    public int Size(int x) => -par[this[x]];

    /// <summary>
    /// 指定した 2 つの要素が属する集合が同じかどうかを判定する．計算量 O(α(N))．
    /// </summary>
    /// <param name="x">最初の要素の 0-indexed での番号</param>
    /// <param name="y">2 つ目の要素の 0-indexed での番号</param>
    /// <returns>同じならば true，そうでなければ false．</returns>
    /// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
    public bool Same(int x, int y) => this[x] == this[y];
}

static class Ex
{
    public static void join<T>(this IEnumerable<T> values, string sep = " ") => WriteLine(string.Join(sep, values));
    public static string concat<T>(this IEnumerable<T> values) => string.Concat(values);
    public static string reverse(this string s) { var t = s.ToCharArray(); Array.Reverse(t); return t.concat(); }

    public static int lower_bound<T>(this IList<T> arr, T val) where T : IComparable<T>
    {
        int low = 0, high = arr.Count;
        int mid;
        while (low < high)
        {
            mid = ((high - low) >> 1) + low;
            if (arr[mid].CompareTo(val) < 0) low = mid + 1;
            else high = mid;
        }
        return low;
    }
    public static int upper_bound<T>(this IList<T> arr, T val) where T : IComparable<T>
    {
        int low = 0, high = arr.Count;
        int mid;
        while (low < high)
        {
            mid = ((high - low) >> 1) + low;
            if (arr[mid].CompareTo(val) <= 0) low = mid + 1;
            else high = mid;
        }
        return low;
    }
}

class Pair<T, U> : IComparable<Pair<T, U>> where T : IComparable<T> where U : IComparable<U>
{
    public T f; public U s;
    public Pair(T f, U s) { this.f = f; this.s = s; }
    public int CompareTo(Pair<T, U> a) => f.CompareTo(a.f) != 0 ? f.CompareTo(a.f) : s.CompareTo(a.s);
    public override string ToString() => $"{f} {s}";
}

class Scanner
{
    string[] s; int i;
    readonly char[] cs = new char[] { ' ' };
    public Scanner() { s = new string[0]; i = 0; }
    public string[] scan => ReadLine().Split();
    public int[] scanint => Array.ConvertAll(scan, int.Parse);
    public long[] scanlong => Array.ConvertAll(scan, long.Parse);
    public double[] scandouble => Array.ConvertAll(scan, double.Parse);
    public string next
    {
        get
        {
            if (i < s.Length) return s[i++];
            string st = ReadLine();
            while (st == "") st = ReadLine();
            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            i = 0;
            return next;
        }
    }
    public int nextint => int.Parse(next);
    public long nextlong => long.Parse(next);
    public double nextdouble => double.Parse(next);
}