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 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
  public static void main(String... args) {
    Scanner input = new Scanner(System.in);
    
    int n = input.nextInt();
    int m = input.nextInt();
    
    Map<Integer, Set<Integer>> g = new HashMap<>();
    for (int i = 0; i < m; i++) {
      int a = input.nextInt() - 1;
      int b = input.nextInt() - 1;
      
      g.computeIfAbsent(a, k -> new HashSet<>()).add(b);
      g.computeIfAbsent(b, k -> new HashSet<>()).add(a);
    }
    
    int ans = 0;
    boolean[] v = new boolean[n];
	for (int i = 0; i < n; i++) {
      if (!v[i]) {
        ans = Math.max(ans, dfs(i, v, g));
      }
    }
    
    System.out.println(ans);
  }
  
  private static int dfs(int root, boolean[] v, Map<Integer, Set<Integer>> g) {
    v[root] = true;
    
    int cnt = 1;
    for (int n : g.getOrDefault(root, new HashSet<>())) {
      if (!v[n]) {
        cnt += dfs(n, v, g);
      }
    }
    
    return cnt;
  }
}