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.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ABC177D
{
    class Program
    {
        static int[] renketu;
        static void Main(string[] args)
        {
            string[] NM = Console.ReadLine().Split(' ');
            int N = int.Parse(NM[0]);
            int M = int.Parse(NM[1]);
            int[] A = new int[M];
            int[] B = new int[M];
            for(int i = 0; i < M; i++)
            {
                string[] AB = Console.ReadLine().Split(' ');
                A[i] = int.Parse(AB[0]);
                B[i] = int.Parse(AB[1]);
            }
            renketu = new int[N];
            for(int i = 0; i < M; i++)
            {
                if (renketu[A[i]-1] == 0 && renketu[B[i]-1] == 0)
                {
                    renketu[A[i] - 1] = -1;
                    renketu[B[i] - 1] = A[i];
                }
                else if (renketu[A[i]-1] != 0 && renketu[B[i]-1] == 0)
                {
                    renketu[B[i] - 1] = A[i];
                    int p = oya(B[i]);
                    renketu[p - 1] -= 1;
                }
                else if (renketu[A[i]-1] == 0 && renketu[B[i]-1] != 0)
                {
                    renketu[A[i] - 1] = B[i];
                    int p = oya(B[i]);
                    renketu[p - 1] -= 1;
                }
                else
                {
                    int p1 = oya(A[i]);
                    int p2 = oya(B[i]);
                    if (p1 != p2)
                    {
                        renketu[p1 - 1] += renketu[p2 - 1] - 1;
                        renketu[p2 - 1] = p1;
                    }
                }
            }
            int max = 0;
            for(int i = 0; i < N; i++)
            {
                if (renketu[i] < 0)
                {
                    int temp = -(renketu[i] - 1);
                    if (temp > max)
                    {
                        max = temp;
                    }
                }
            }
            Console.WriteLine(max);
        }

        static int oya(int a)
        {
            if (renketu[a - 1] > 0)
            {
                renketu[a - 1] = oya(renketu[a - 1]);
                return renketu[a - 1];
            }
            return a;
        }
    }
}