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: Runtime Error)

import java.util.*; 
import java.lang.*; 


public class Main{
	public static void main(String[] args){
		Scanner scan = new Scanner(System.in);
                
                int N = scan.nextInt();
                int M = scan.nextInt();
                
                int[] A = new int[M+1];
                int[] B = new int[M+1];
                
                int[][] cnt = new int[N+1][N+1];
                int[] ans = new int[N+1];
                
                for(int i=1; i<=M;i++){
                    A[i] = scan.nextInt();
                    B[i] = scan.nextInt();
                    
                    if(cnt[A[i]][B[i]]!=1){
                        cnt[A[i]][B[i]]++;
                        ans[A[i]]++;
                    }
                    if(cnt[B[i]][A[i]]!=1){
                        cnt[B[i]][A[i]]++;
                        ans[B[i]]++;
                    }
                }
                int k = 0;
                for(int i=1; i<=N;i++){
                    k=Math.max(k, ans[i]);
                }
                
                System.out.println(k+1);
        }
}