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: Time Limit Exceeded)

import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class Main{
	private static ArrayList<Integer> UnionFind = new ArrayList<Integer>();
	public static void main(String args[]) {
		Scanner stdIn = new Scanner(System.in);
		int N = stdIn.nextInt();
		int M = stdIn.nextInt();
		// initialize
		for(int i = 0; i < N; i ++) {
			UnionFind.add(-1);
		}
		for(int i = 0; i < M; i ++) {
			unit(stdIn.nextInt() - 1, stdIn.nextInt() - 1);
		}
		System.out.println(Collections.min(UnionFind) * -1);
	}
	
	private static int getRoot(int x) {
		if(UnionFind.get(x) < 0) {
			return x;
		}else {
			return getRoot(UnionFind.get(x));	
		}
	}
	
	private static void unit(int A, int B) {
		int rootA = getRoot(A);
		int rootB = getRoot(B);
		if(rootA != rootB) {
			UnionFind.set(rootA, UnionFind.get(rootA) + UnionFind.get(rootB));
			UnionFind.set(rootB, rootA);
		}
	}
}