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;

namespace AtCoderBeginnerContest177
{
	class Program
	{
		static void Main(string[] args)
		{
			var inputs = Console.ReadLine().Split().Select(int.Parse).ToList();

			if (inputs[1] == 0)
			{
				Console.WriteLine(1);
				return;
			}

			var relation = new List<(int, int)>();
			for (var i = 0; i < inputs[1]; i++)
			{
				var tmp = Console.ReadLine().Split().Select(int.Parse).OrderBy(x => x).ToList();
				relation.Add((tmp[0], tmp[1]));
			}
			relation = relation.OrderBy(x => x.Item1).Distinct().ToList();

			var groups = new List<int>();
			//var groupIncludes = new List<bool[]>();
			var allIncludes = new bool[inputs[0]];
			var groupMap = new int[inputs[0]];
			for (var i = 0; i < relation.Count; i++)
			{
				if (allIncludes[relation[i].Item1 - 1] && allIncludes[relation[i].Item2 - 1])
				{
					if (groupMap[relation[i].Item1 - 1] != groupMap[relation[i].Item2 - 1])
					{
						groups[relation[i].Item1 - 1] += groups[relation[i].Item2 - 1];
						var target = groupMap[relation[i].Item2 - 1];
						for (var j = 0; j < groupMap.Length; j++)
						{
							if (groupMap[j] == target)
							{
								groupMap[j] = groupMap[relation[i].Item1 - 1];
							}
						}
					}
					continue;
				}

				if (!allIncludes[relation[i].Item1 - 1] && !allIncludes[relation[i].Item2 - 1])
				{
					groups.Add(2);
					groupMap[relation[i].Item1 - 1] = groups.Count - 1;
					groupMap[relation[i].Item2 - 1] = groups.Count - 1;
					//var array = new bool[inputs[0]];
					//array[relation[i].Item1 - 1] = true;
					//array[relation[i].Item2 - 1] = true;
					allIncludes[relation[i].Item1 - 1] = true;
					allIncludes[relation[i].Item2 - 1] = true;
					//groupIncludes.Add(array);
				}
				else
				{
					if (allIncludes[relation[i].Item1 - 1])
					{
						groups[groupMap[relation[i].Item1 - 1]]++;
						groupMap[relation[i].Item2 - 1] = groupMap[relation[i].Item1 - 1];
						//groupIncludes[groupMap[relation[i].Item1 - 1]][relation[i].Item2 - 1] = true;
						allIncludes[relation[i].Item2 - 1] = true;
					}
					else
					{
						groups[groupMap[relation[i].Item2 - 1]]++;
						groupMap[relation[i].Item1 - 1] = groupMap[relation[i].Item2 - 1];
						//groupIncludes[groupMap[relation[i].Item2 - 1]][relation[i].Item1 - 1] = true;
						allIncludes[relation[i].Item1 - 1] = true;
					}
				}
			}
			Console.WriteLine(groups.Max());
		}
	}
}