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: Wrong Answer)

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

namespace AtCoder.D
{
    public class Program
    {
        public static void Main() { var r = GetResult(); Debug.WriteLine(r); Console.Write(r); }

        private static object GetResult()
        {
            var (N, M) = Read2Longs();
            var group = new int[N];
            var max = 1;
            for (var i = 0; i < M; i++)
            {
                var (A, B) = Read2Longs();
                if (group[A - 1] == 0 && group[B - 1] == 0)
                {
                    group[A - 1] = max;
                    group[B - 1] = max;
                    max++;
                }
                else if(group[A-1] != 0)
                {
                    group[B - 1] = group[A - 1];
                }
                else
                {
                    group[A - 1] = group[B - 1];
                }
            }
            return group.GroupBy(x => x).Select(x => x.Count()).Max();
        }

        #region Console.Read

        public static string ReadText() { return Console.ReadLine(); }
        public static List<string> ReadTexts() { return Console.ReadLine().Split(' ').ToList(); }
        public static long ReadLong() { return long.Parse(Console.ReadLine()); }
        public static (long, long) Read2Longs() { var ls = ReadLongs(); return (ls[0], ls[1]); }
        public static (long, long, long) Read3Longs() { var ls = ReadLongs(); return (ls[0], ls[1], ls[2]); }
        public static (long, long, long, long) Read4Longs() { var ls = ReadLongs(); return (ls[0], ls[1], ls[2], ls[3]); }
        public static (long, long, long, long, long) Read5Longs() { var ls = ReadLongs(); return (ls[0], ls[1], ls[2], ls[3], ls[4]); }
        public static List<long> ReadLongs() { return Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToList(); }

        #endregion
    }
}