Problem:
Score : 400 points

Problem Statement

There are N persons called Person 1 through Person N.

You are given M facts that "Person A_i and Person B_i are friends." The same fact may be given multiple times.

If X and Y are friends, and Y and Z are friends, then X and Z are also friends. There is no friendship that cannot be derived from the M given facts.

Takahashi the evil wants to divide the N persons into some number of groups so that every person has no friend in his/her group.

At least how many groups does he need to make?

Constraints

2 \leq N \leq 2\times 10^5

0 \leq M \leq 2\times 10^5

1\leq A_i,B_i\leq N

A_i \neq B_i

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
\vdots
A_M B_M

Output

Print the answer.

Sample Input 1

5 3
1 2
3 4
5 1

Sample Output 1

3

Dividing them into three groups such as \{1,3\}, \{2,4\}, and \{5\} achieves the goal.

Sample Input 2

4 10
1 2
2 1
1 2
2 1
1 2
1 3
1 4
2 3
2 4
3 4

Sample Output 2

4

Sample Input 3

10 4
3 1
4 1
5 9
2 6

Sample Output 3

3

Solution in C# (.NET Core 3.1.201) (judge verdict: Time Limit Exceeded)

using System;
using System.Linq;
using System.Collections.Generic;
using System.Text.RegularExpressions;

namespace AtCoderBeginnerContest177
{
    class Program
    {
        static void Main(string[] args)
        {

            // 課題D：Friends(2020/8/30)

            string[] str = Console.ReadLine().Split(' ');
            int rows = int.Parse(str[1]);

            if (rows == 0)
            {
                Console.Write(1);
                return;
            }

            //HashSet<int[]> relation = new HashSet<int[]>();

            int[] pair;
            HashSet<int[]> groupArray = new HashSet<int[]>();

            HashSet<int> groupMember = new HashSet<int>();
            HashSet<int> tmpGroupMember = new HashSet<int>();
            int arraySize = 2;
            bool gFlg;

            int maxGroup = 0;

            for (int i = 0; i < rows; i++)
            {
                str = Console.ReadLine().Split(' ');
                pair = new int[] { int.Parse(str[0]), int.Parse(str[1]) };

                //if (relation.Contains(pair)) continue;
                //relation.Add(pair);

                gFlg = false;

                groupMember.Clear();

                foreach (int[] group in groupArray)
                {
                    
                    if (!group.Contains(pair[0]) && !group.Contains(pair[1])) continue;

                    tmpGroupMember.Clear();

                    tmpGroupMember.UnionWith(group);
                    tmpGroupMember.UnionWith(pair);

                    gFlg = true;

                    groupMember.UnionWith(tmpGroupMember);

                }

                if (!gFlg)
                {
                    groupArray.Add(pair);
                }
                else
                {
                    arraySize = groupMember.Count();
                    pair = new int[arraySize];
                    groupMember.CopyTo(pair);

                    groupArray.Add(pair);
                }
                if (maxGroup < arraySize) maxGroup = arraySize;
            }

            Console.Write(maxGroup);

        }
    }
}