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[] tmpSet;

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

                if (!relation.Contains(tmpSet)) relation.Add(tmpSet);

            }

            List<int[]> groupArray = new List<int[]>();

            HashSet<int> groupMember = new HashSet<int>();
            HashSet<int> tmpGroupMember = new HashSet<int>();
            int[] pair;
            int p1, p2, groupNum;
            bool gFlg;

            foreach (int[] friends in relation)
            {
                p1 = friends[0];
                p2 = friends[1];

                gFlg = false;

                groupMember.Clear();

                foreach (int[] group in groupArray)
                {
                    tmpGroupMember.Clear();

                    for (int j = 0; j < group.Count(); j++)
                    {
                        int tmpIntger = group[j];
                        tmpGroupMember.Add(tmpIntger);
                    }

                    if (!tmpGroupMember.Contains(p1) && !tmpGroupMember.Contains(p2)) continue;

                    tmpGroupMember.Add(p1);
                    tmpGroupMember.Add(p2);

                    if (!gFlg) gFlg = true;

                    foreach (int p in tmpGroupMember) groupMember.Add(p);

                }

                if (!gFlg)
                {
                    pair = new int[]{ p1, p2 };
                    groupArray.Add(pair);
                }
                else
                {
                    int arraySize = groupMember.Count();
                    int arrayCnt = 0;
                    pair = new int[arraySize];
                    foreach (int p in groupMember)
                    {
                        pair[arrayCnt] = p;
                        arrayCnt++;
                    }
                    groupArray.Add(pair);
                }
            }

            int maxGroup = 0;

            for (int i = 0; i < groupArray.Count; i++)
            {
                int cntPeople = groupArray[i].Count();

                if (maxGroup < cntPeople) maxGroup = cntPeople;

            }

            Console.Write(maxGroup);

        }
    }
}