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.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Text;

class Program
{
    private static int depth = 0;

    static void Main(string[] args)
    {
        var input = Console.ReadLine().Split(" ");
        var N = int.Parse(input[0]);
        var M = int.Parse(input[1]);

        var graph = new List<List<int>>();
        var seen = new List<bool>();
        graph.Add(new List<int>());
        seen.Add(true);
        for (int i = 0; i < N; i++)
        {
            graph.Add(new List<int>());
            seen.Add(false);
        }

        for (int i = 0; i < M; i++)
        {
            var input2 = Console.ReadLine().Split(" ");
            var A = int.Parse(input2[0]);
            var B = int.Parse(input2[1]);

            graph[A].Add(B);
            graph[B].Add(A);
        }

        var depthMax = 0;
        for (int i = 1; i < N; i++)
        {
            if (seen[i]) continue;
            depth = 0;
            dfs(graph, seen, i);
            depthMax = Math.Max(depth, depthMax);
        }

        Console.WriteLine(depthMax);
    }

    private static void dfs(List<List<int>> graph, List<bool> seen, int v)
    {
        depth++;
        // vを訪問済みにする
        seen[v] = true;

        // vから行ける各頂点next_vについて再帰処理
        foreach (var next_v in graph[v])
        {
            if (seen[next_v]) continue; //next_vが探索済みならスルー
            dfs(graph, seen, next_v);
        }
    }
}