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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.LinkedList;

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		String[] tokens = in.readLine().split(" ");
		int n = Integer.parseInt(tokens[0]);
		int m = Integer.parseInt(tokens[1]);
		ArrayList<Integer>[] adj = new ArrayList[n+1];
		for (int i = 1; i <= n; i++)
			adj[i] = new ArrayList<Integer>();
		int a, b;
		for (int i = 0; i < m; i++) {
			tokens = in.readLine().split(" ");
			a = Integer.parseInt(tokens[0]);
			b = Integer.parseInt(tokens[1]);
			adj[a].add(b);
			adj[b].add(a);
		}
		boolean[] v = new boolean[n+1];
		int ans = 0;
		int size;
		for (int i = 1; i <= n; i++) {
			if (!v[i]) {
				LinkedList<Integer> q = new LinkedList<Integer>();
				q.add(i);
				v[i] = true;
				size = 1;
				while (!q.isEmpty()) {
					int c = q.poll();
					for (int j : adj[c]) {
						if (!v[j]) {
							q.add(j);
							v[j] = true;
							size++;
						}
					}
				}
				ans = Math.max(size, ans);
			}
		}
		System.out.println(ans);
	}
}