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: Wrong Answer)

import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        Map<Integer, Vertex> vMap = new HashMap<>();
        for (int i = 0; i < m; i++) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            if (!vMap.containsKey(a)) {
                vMap.put(a, new Vertex(a));
            }
            if (!vMap.containsKey(b)) {
                vMap.put(b, new Vertex(b));
            }
            vMap.get(a).nbrs.add(vMap.get(b));
            vMap.get(b).nbrs.add(vMap.get(a));
        }

        final Set<Integer> vis = new HashSet<>();
        int ans = 0;
        int lastSize = 0;
        for (int i = 1; i <= n; i++) {
            if (!vis.contains(i) && vMap.containsKey(i)) {
                bfs(vMap, i, vis);
                ans = Math.max(ans, vis.size() - lastSize);
                lastSize = vis.size();
            }
        }
        System.out.println(ans);
    }

    private static void bfs(Map<Integer, Vertex> vMap, int v, Set<Integer> visited) {
        Queue<Integer> q = new LinkedList<>();
        q.add(v);
        while (!q.isEmpty()) {
            int x = q.poll();
            for (Vertex n : vMap.get(x).nbrs) {
                if (!visited.contains(n.id)) {
                    q.add(n.id);
                    visited.add(n.id);
                }
            }
        }
    }

    static class Vertex {
        int id;
        List<Vertex> nbrs = new ArrayList<>();

        public Vertex(int id) {
            this.id = id;
        }
    }
}