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: Accepted)



import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Set;
import java.util.StringTokenizer;

public class Main {

	public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());

        HashMap<Integer, Set<Integer>> map = new HashMap<>();
        for(int i=1; i<=n; i++){
            map.put(i, new HashSet<Integer>());
        }

        for(int i=0; i<m; i++){
            StringTokenizer pair = new StringTokenizer(br.readLine());
            int v1 = Integer.parseInt(pair.nextToken());
            int v2 = Integer.parseInt(pair.nextToken());

            map.get(v1).add(v2);
            map.get(v2).add(v1);
        }

        int max = 0;
        boolean[] vis = new boolean[n+1];

        for(int i=1; i<=n; i++){
            if(!vis[i]){
                max = Math.max(max, dfs(i, vis, map));
            }
        }

        System.out.println(max);
    }

    static int dfs(int i, boolean[] vis, HashMap<Integer, Set<Integer>> map){
        vis[i] = true;
        int size = 1;

        Set<Integer> fr = map.get(i);
        
        for(int next: fr){
            if(!vis[next]){
                size += dfs(next, vis, map);
            }
        }

        return size;
    }
}