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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.*;
import java.util.*;

class Main
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        String s[]=bu.readLine().split(" ");
        int n=Integer.parseInt(s[0]),m=Integer.parseInt(s[1]);
        int i,max=0,c=0;
        ArrayList<Integer> g[]=new ArrayList[n];
        boolean vis[]=new boolean[n];
        for(i=0;i<n;i++)
        g[i]=new ArrayList<>();
        for(i=0;i<m;i++)
        {
            s=bu.readLine().split(" ");
            int u=Integer.parseInt(s[0])-1,v=Integer.parseInt(s[1])-1;
            g[u].add(v);
            g[v].add(u);
        }

        for(i=0;i<n;i++)
        if(!vis[i])
        {
            Queue<Integer> q=new LinkedList<>();
            vis[i]=true;
            q.add(i);
            c=0;
            while(!q.isEmpty())
            {
                int p=q.poll();
                c++;
                for(int x:g[p])
                if(!vis[x])
                {
                    vis[x]=true;
                    q.add(x);
                }
            }
            max=Math.max(max,c);
        }
        System.out.print(max);
    }
}