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.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {


		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int m = sc.nextInt();

		ArrayList<HashSet<Integer>> group = new ArrayList<HashSet<Integer>>();
		HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
		int index=0;
		for (int i=0; i<m; i++){
			int a = sc.nextInt();
			int b = sc.nextInt();
			if (map.containsKey(a) || map.containsKey(b)){
				int indexA = 0;
				int indexB = 0;
				int index2 = 0;
				if (map.containsKey(a)){
					indexA = map.get(a);
					index2 = indexA;
				}
				if (map.containsKey(b)){
					indexB = map.get(b);
					index2 = indexB;
				}
				if ((map.containsKey(a) && map.containsKey(b)) && indexA != indexB){
					HashSet<Integer> setA = group.get(indexA);
					HashSet<Integer> setB = group.get(indexB);
					for (int x : setB){
						setA.add(x);
						map.put(x, indexA);
					}
				} else {
					group.get(index2).add(a);
					group.get(index2).add(b);
					map.put(a, index2);
					map.put(b, index2);
				}
			} else {
				HashSet<Integer> set = new HashSet<Integer>();
				set.add(a);
				set.add(b);
				group.add(set);
				map.put(a, index);
				map.put(b, index);
				index++;
			}
		}
		int ans=0;
		for (int i=0; i<group.size();i++){
			ans = Math.max(group.get(i).size(), ans);
		}
		System.out.println(ans);
	}


}