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: Runtime Error)

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

namespace learning
{
    class Program
    {
        static void Main()
        {
            // 課題D：Friends(2020/8/30)

            string[] str = Console.ReadLine().Split(' ');
            int people = int.Parse(str[0]);
            int rows = int.Parse(str[1]);

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

            HashSet<string[]> relation = new HashSet<string[]>();
            for (int i = 0; i < rows; i++) relation.Add(Console.ReadLine().Split(' '));

            int arraySize = people+1;

            int[,] groupArray = new int[arraySize, arraySize];
            
            for (int i = 0; i < arraySize; i++)
            {
                for (int j = 0; j < arraySize; j++)
                {
                    groupArray[i, j] = 0;
                }
            }

            int arrayIndex = 0;

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

                int groupNum = -1;
                int cnt = 0;
                HashSet<int> groupMember = new HashSet<int>();

                for (int i = 0; i < arraySize; i++)
                {
                    HashSet<int> tmpGroupMember = new HashSet<int>();

                    for (int j = 0; j < arraySize; j++)
                    {
                        if (groupArray[i, j] == 0) continue;
                        tmpGroupMember.Add(groupArray[i, j]);

                    }

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

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

                    if (groupNum < 0) groupNum = i;
                    
                    for (int j = 0; j < arraySize; j++) groupArray[i, j] = 0;

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

                }

                if (groupNum < 0)
                {
                    groupArray[arrayIndex, 0] = p1;
                    groupArray[arrayIndex, 1] = p2;
                    arrayIndex++;
                }
                else
                {
                    foreach (int p in groupMember)
                    {
                        groupArray[groupNum, cnt] = p;
                        cnt++;
                    }
                }
            }

            int maxGroup = 0;

            for (int i = 0; i < arraySize; i++)
            {
                int cntPeople = 0;

                for (int j = 0; j < arraySize; j++)
                {
                    if (groupArray[i, j] != 0) cntPeople++;

                }

                if (maxGroup < cntPeople) maxGroup = cntPeople;

            }

            Console.Write(maxGroup);
        }
    }
}