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: Wrong Answer)

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

namespace ProgramingStudying2
{
    class Program
    {

        static void Main(string[] args)
        {
            var nm = Console.ReadLine().Split().Select(int.Parse).ToArray();
            var n = nm[0];
            var m = nm[1];

            var G = new List<List<int>>(); 
            for(int i = 0; i < n; i++)
            {
                G.Add(new List<int>());
            }

            for (int i = 0; i < m; i++)
            {
                var ab = Console.ReadLine().Split();
                var a = int.Parse(ab[0]) - 1;
                var b = int.Parse(ab[1]) - 1;

                G[a].Add(b);
                G[b].Add(a);
            }

            var ans = 0;
            var visited = new bool[n];
            var que = new Queue<int>();

            for(int i = 0; i < n; i++)
            {
                if (visited[i])
                {
                    continue;
                }

                var count = 0;

                que.Enqueue(i);
                while (que.Any())
                {
                    var x = que.Dequeue();

                    foreach(var nx in G[x])
                    {
                        if (visited[nx])
                        {
                            continue;
                        }

                        visited[nx] = true;
                        que.Enqueue(nx);
                        count++;
                    }
                }
                if(count > ans)
                {
                    ans = count;
                }
            }
            Console.WriteLine(ans);
        }
    }
}