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: Wrong Answer)

import java.util.*;

public class Main{
  public static void main(String args[]){
    Scanner sc=new Scanner(System.in);
    int N=sc.nextInt();
    int M=sc.nextInt();
    Integer[][] data=new Integer[2][M];
    List<List<Integer>> friends=new ArrayList<>();
    int res1=-1;
	int res2=-1;
    for(int i=0;i<M;i++){
	 data[0][i]=sc.nextInt();
     data[1][i]=sc.nextInt();
	 res1=res2=-1;     
     for(int j=0;j<friends.size();j++){
       if(friends.get(j).contains(data[0][i])){
         res1=j;
       }
       if(friends.get(j).contains(data[1][i])){
         res2=j;
       }
     }
     if(res1!=-1 && res2==-1){
       friends.get(res1).add(data[1][i]);
     }else if(res1==-1 && res2!=-1){
       friends.get(res2).add(data[0][i]);
     }else if(res1!=-1 && res2!=-1){
       friends.get(res1).addAll(friends.get(res2));
       friends.remove(res2);
     }else{
		List c=new ArrayList<Integer>();
       c.add(data[0][i]);
       c.add(data[1][i]);
       friends.add(c);
              
     }
    }
    int biggest=0;
	for(int i=0;i<friends.size();i++){
      if(biggest<friends.get(i).size()){
        biggest=friends.get(i).size();
      }
    }
    System.out.println(biggest);
  }
}