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.Linq;
using System.Collections.Generic;
using System.Numerics;
namespace ABC177
{
    class Program
    {
        static void Main(string[] args)
        {
            var NM = Console.ReadLine().Split().Select(long.Parse).ToArray();
            var N = NM[0]; var M = NM[1];
            var CA = Enumerable.Repeat(0, (int)M).Select(_ => Console.ReadLine().Split().Select(long.Parse).ToArray()).OrderBy(x => x[0]).ToArray();

            var group = new long[N];
            var gnum = 1L;

            // Console.WriteLine(CA[0][0]);

            for (int i = 0; i < M; i++)
            {
                if (group[CA[i][0] - 1] == 0 && group[CA[i][1] - 1] == 0)
                {
                    group[CA[i][0] - 1] = gnum;
                    group[CA[i][1] - 1] = gnum;
                    gnum++;
                }
                else
                {
                    if (group[CA[i][0] - 1] == 0) group[CA[i][0] - 1] = group[CA[i][1] - 1];
                    if (group[CA[i][1] - 1] == 0) group[CA[i][1] - 1] = group[CA[i][0] - 1];
                }
            }
            //Console.WriteLine(string.Join(" ", group));

            var gcount = new long[N];
            for (int i = 0; i < N; i++)
            {
                if (group[i] != 0) gcount[group[i] - 1]++;
            }

            gcount = gcount.OrderBy(x => x).ToArray();

            //Console.WriteLine(string.Join(" ", gcount));
            var sum = 0L;
            var isFirst = true;
            for (int i = 0; i < N; i++)
            {
                if (gcount[i] == 0) continue;
                if (isFirst) { sum += gcount[i]; isFirst = false; continue; }
                sum += gcount[i] - sum;
            }

            //Console.WriteLine(string.Join(" ", group));
            if (sum == 0) sum = 1;
            Console.WriteLine(sum);

        }
    }
}