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: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using static System.Console;

class Program
{
    static void Main(string[] args)
    {
        var NM = ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
        var (N, M) = (NM[0], NM[1]);
        var AB = Enumerable.Range(0, M).Select(i => ReadLine().Split(' ')).Select(s => (A: int.Parse(s[0]) - 1, B: int.Parse(s[1]) - 1)).ToArray();

        var adjacencyList = new HashSet<int>[N];
        for (int i = 0; i < adjacencyList.Length; i++)
            adjacencyList[i] = new HashSet<int>();

        for (int i = 0; i < M; i++)
        {
            var (A, B) = AB[i];
            adjacencyList[A].Add(B);
            adjacencyList[B].Add(A);
        }

        var gCount = 0;

        var visited = new int[N];
        for (int i = 0; i < visited.Length; i++)
            visited[i] = -1;

        for (int i = 0; i < N; i++)
        {
            if (visited[i] != -1)
                continue;

            gCount++;

            var q = new Queue<int>();

            var v = i;
            visited[v] = gCount;
            q.Enqueue(v);

            while (0 < q.Count)
            {
                v = q.Dequeue();

                foreach (var next in adjacencyList[v])
                {
                    if (visited[next] == -1)
                    {
                        visited[next] = gCount;
                        q.Enqueue(next);
                    }
                }
            }
        }

        var ans = visited.GroupBy(x => x).Select(g => g.Count()).Max();

        WriteLine(ans);
    }
}