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: Runtime Error)

import java.util.Scanner;
import java.util.Queue;
import java.util.ArrayDeque;
 
public class Main {
	static int cnt = 0;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		int m = sc.nextInt();
		int[] a = new int[m];
		int[] b = new int[m];
		boolean[] searched = new boolean[n+1];
		for(int i= 0; i < m; i++) {
			a[i] = sc.nextInt();
			b[i] = sc.nextInt();
		}
		for(int i = 0; i < n + 1; i++) {
			searched[i] = false;
		}
		int ans = 0;
		Queue<Integer> friend = new ArrayDeque<>();
		friend.add(a[0]);
		
		label: while(true) {
			//System.out.println(1);
			cnt = 0;
			search(friend.poll(), a, b, searched, friend);
			//System.out.println(cnt);
			if(ans < cnt) {
				ans = cnt;
			}
			for(int i = 0; i < m; ++i) {
				//System.out.println(i);
				//System.out.println((i==(m-1))+" "+i+" "+m);
				if(!searched[a[i]]) {
					//System.out.println(searched[a[i]]);
					friend.add(a[i]);
					continue label;
				}else if(!searched[b[i]]) {
					//System.out.println(searched[b[i]]);
					friend.add(b[i]);
					continue label;
				}else if(i == (m - 1)) {
					break label;
				}
			}
		}
		System.out.println(ans);
	}
		
	public static void search(int root,int[] a,int[] b,boolean[] searched,Queue<Integer> friend) {
		//System.out.println(root);
		searched[root] = true;
		++cnt;
		for(int i = 0; i < a.length; ++i) {
			if(a[i] == root && searched[b[i]] == false) {
				friend.add(b[i]);
				searched[b[i]] = true;
			}else if(b[i] == root && searched[a[i]] == false) {
				friend.add(a[i]);
				searched[a[i]] = true;
			}
		}
		if(friend.peek() != null) {
			search(friend.poll(),a,b,searched,friend);
		}
	}
}