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.*;
 
public class Main {
	public static void main (String[] args) {
	  Scanner sc = new Scanner(System.in);
      int n = sc.nextInt();
      int m = sc.nextInt();
      int[] rank = new int[n];
      int[] parent = new int[n];
      
      for(int i= 0; i<n; i++){
        rank[i] = 1;
        parent[i] = i;
      }
      
      int res = 0;
      for(int i=0; i<m; i++){
          int u = sc.nextInt();
          int v = sc.nextInt();	
          union(u, v, rank, parent);
      }
      for(int curRank : rank) res = Math.max(res, curRank);
      System.out.println(res);
	}
  public static int find(int u, int[] parent){
    if(u != parent[u]){
      parent[u] = find(parent[u], parent);
    }
    return parent[u];
  }
  
  public static void union(int u, int v, int[] rank, int[] parent){
    int parU = find(u, parent);
    int parV = find(v, parent);
    if(parV == parU) return;
    if(rank[parV] > rank[parU]){
      rank[parV] += rank[parU];
      parent[parU] = parV;
      return;
    }
    rank[parU] += rank[parV];
    parent[parV] = parU;
    return;
  }
}