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# (Mono-csc 3.5.0) (judge verdict: Runtime Error)

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

class Program
{
    static string InputPattern = "InputX";

    static List<string> GetInputList()
    {
        var WillReturn = new List<string>();

        if (InputPattern == "Input1") {
            WillReturn.Add("5 3");
            WillReturn.Add("1 2");
            WillReturn.Add("3 4");
            WillReturn.Add("5 1");
            //3
        }
        else if (InputPattern == "Input2") {
            WillReturn.Add("4 10");
            WillReturn.Add("1 2");
            WillReturn.Add("2 1");
            WillReturn.Add("1 2");
            WillReturn.Add("2 1");
            WillReturn.Add("1 2");
            WillReturn.Add("1 3");
            WillReturn.Add("1 4");
            WillReturn.Add("2 3");
            WillReturn.Add("2 4");
            WillReturn.Add("3 4");
            //4
        }
        else if (InputPattern == "Input3") {
            WillReturn.Add("10 4");
            WillReturn.Add("3 1");
            WillReturn.Add("4 1");
            WillReturn.Add("5 9");
            WillReturn.Add("2 6");
            //3
        }
        else {
            string wkStr;
            while ((wkStr = Console.ReadLine()) != null) WillReturn.Add(wkStr);
        }
        return WillReturn;
    }

    static void Main()
    {
        List<string> InputList = GetInputList();

        int[] wkArr = { };
        Action<string> SplitAct = pStr =>
            wkArr = pStr.Split(' ').Select(X => int.Parse(X)).ToArray();

        foreach (string EachStr in InputList.Skip(1)) {
            SplitAct(EachStr);
            RenketuSyori(wkArr[0], wkArr[1]);
        }
        Console.WriteLine(UnionSetList.Max(pX => pX.Count));
    }

    //UnionFindのSetのList
    static List<HashSet<int>> UnionSetList = new List<HashSet<int>>();

    //連結処理を行う
    static void RenketuSyori(int p1, int p2)
    {
        //同じノードなら何もしない
        if (p1 == p2) return;

        int wkInd1 = UnionSetList.FindIndex(X => X.Contains(p1));
        int wkInd2 = UnionSetList.FindIndex(X => X.Contains(p2));

        //UnionFind木に両方とも存在しない場合
        if (wkInd1 < 0 && wkInd2 < 0) {
            UnionSetList.Add(new HashSet<int>() { p1, p2 });
        }
        //UnionFind木にp2のみが存在する場合
        else if (wkInd1 < 0 && wkInd2 >= 0) {
            UnionSetList[wkInd2].Add(p1);
        }
        //UnionFind木にp1のみが存在する場合
        else if (wkInd1 >= 0 && wkInd2 < 0) {
            UnionSetList[wkInd1].Add(p2);
        }
        //別々のUnionFind木なら木をマージ
        else if (wkInd1 != wkInd2) {
            UnionSetList[wkInd1].UnionWith(UnionSetList[wkInd2]);
            UnionSetList.RemoveAt(wkInd2);
        }
    }
}