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.Collections.Generic;
using System.Linq;

class Program
{
  static void Main(string[] args)
  {
    string[] firstLine = Console.ReadLine().Split(' ');
    uint N = uint.Parse(firstLine[0]);
    uint M = uint.Parse(firstLine[1]);

    uint[,] friendMap = new uint[N, N];

    for (int i = 0; i < M; i++)
    {
      string[] items = Console.ReadLine().Split(' ');
      uint A = uint.Parse(items[0]) - 1;
      uint B = uint.Parse(items[1]) - 1;
      if (A < B)
      {
        friendMap[A, B] = 1;
      }
      else
      {
        friendMap[B, A] = 1;
      }
    }

    uint[] groupNum = new uint[N];
    for (uint i = 0; i < N; i++)
    {
      for (uint j = i + 1; j < N; j++)
      {
        if (friendMap[i, j] == 1)
        {
          groupNum[i] += GetGroupNum(N, friendMap, j, i);
          groupNum[i] += 1;
        }
      }

      groupNum[i] += 1;
    }

    Console.WriteLine(groupNum.Max());
    return;
  }

  static uint GetGroupNum(uint N, uint[,] friendMap, uint target, uint root)
  {
    uint groupNum = 0;
    for (uint j = target + 1; j < N; j++)
    {
      if (friendMap[target, j] == 1 && friendMap[root, j] != 1)
      {
        groupNum += GetGroupNum(N, friendMap, j, root);
        groupNum += 1;
      }
    }
    return groupNum;
  }
}