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;

namespace ConsoleApp1
{
    class ABC117D
    {
        // https://atcoder.jp/contests/abc177/tasks/abc177_d

        static int[] ls;

        static void Main(string[] args)
        {
            var NM = Console.ReadLine().Split();
            int N = Int32.Parse(NM[0]);
            int M = Int32.Parse(NM[1]);

            ls = new int[N];
            for (int i = 0; i < N; i++)
            {
                ls[i] = -1;
            }

            for (int i = 0; i < M; i++)
            {
                var AB = Console.ReadLine().Split();
                int A = Int32.Parse(AB[0]) - 1;
                int B = Int32.Parse(AB[1]) - 1;

                bool a = hasFriend(A);
                bool b = hasFriend(B);

                int C = A;
                int D = B;

                if (a && b)
                {
                    C = getParent(A);
                    D = getParent(B);

                    if (C == D)
                    {
                        continue;
                    }
                }
                else if (a)
                {
                    C = getParent(A);
                }
                else if (b)
                {
                    C = getParent(B);
                    D = A;
                }

                ls[C] += ls[D]; // 人数加算
                ls[D] = C; // setParent
            }

            int max = 1;
            for (int i = 0; i < N; i++)
            {
                max = Math.Max(max, -ls[i]);
            }

            Console.WriteLine(max);
        }

        static int getParent(int A)
        {
            if (ls[A] > 0)
            {
                return getParent(ls[A]);
            }
            return A;
        }

        static bool hasFriend(int A)
        {
            return (ls[A] != -1);
        }
    }
}