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){
        Main main = new Main();
        main.input();
        main.solve();
        main.output();
    }

    int N; // 2 <= N <= 200000
    int M; // 0 <= M <= 200000
    int[] A;
    int[] B;
    Set<Integer>[] groups;
    int result;

    private void input() {
        Scanner sc = new Scanner(System.in);
        N = sc.nextInt();
        M = sc.nextInt();
        A = new int[M];
        B = new int[M];
        groups = new Set[N];
        for (int i=0; i<M; i++) {
            A[i] = sc.nextInt();
            B[i] = sc.nextInt();
        }
        for (int i=0; i<N; i++) {
            Set<Integer> s = new HashSet<>();
            s.add(i+1);
            groups[i] = s;
        }
        result = 1;
    }

    private void solve() {
        for (int i=0; i<M; i++) {
            // debug();
            Set<Integer> groupA = groups[A[i]-1];
            Set<Integer> groupB = groups[B[i]-1];
            Set<Integer> groupAandB = merge(groupA, groupB);
            if (groupAandB.size() > result) {
                result = groupAandB.size();
            }
            groups[A[i]-1] = groupAandB;
            groups[B[i]-1] = groupAandB;
        }
    }

    private Set<Integer> merge(Set<Integer> s1, Set<Integer> s2) {
        Set<Integer> result = new HashSet<>();
        result.addAll(s1);
        result.addAll(s2);
        return result;
    }

    private void output() {
        System.out.println(result);
    }
    
    private void debug() {
        System.out.println("result=" + result);
        for (Set<Integer> g : groups) {
            System.out.println("group_start");
            for (Integer i : g) {
                System.out.println(i);
            }
            System.out.println("group_end");
        }
    }

}