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: Runtime Error)


import java.util.HashMap;
import java.util.Map;

public class Main {

	public static void main(String[] args) {
		var sc = new java.util.Scanner(System.in);
		var N = sc.nextInt();
		var M = sc.nextInt();
		var A = new Integer[M];
		var B = new Integer[M];
		for (int i = 0; i < M; i++) {
			A[i] = sc.nextInt() - 1;
			B[i] = sc.nextInt() - 1;
		}
		Map<Integer, Integer> grpCnt = new HashMap<>();

		var grpid = 1;
		var belongs = new Integer[N];

		for (int i = 0; i < M; i++) {
			if ((belongs[A[i]] == null) && (belongs[B[i]] == null)) {
				belongs[A[i]] = grpid;
				belongs[B[i]] = grpid;
				grpCnt.put(grpid, 2);
				grpid++;
			} else {
				if (belongs[A[i]] == null || belongs[B[i]] == null) {
					if (belongs[A[i]] != null) {
						belongs[B[i]] = belongs[A[i]];
					} else {
						belongs[A[i]] = belongs[B[i]];
					}
					var cnt = grpCnt.get(belongs[A[i]]);
					grpCnt.put(belongs[A[i]], cnt + 1);
				} else if (belongs[A[i]] != belongs[B[i]]) {
					var livingId = Math.max(belongs[A[i]], belongs[B[i]]);
					var deathId = Math.min(belongs[A[i]], belongs[B[i]]);
					var deathCnt = grpCnt.get(deathId);
					var livingCnt = grpCnt.get(livingId);
					grpCnt.remove(deathId);
					grpCnt.put(livingId, deathCnt + livingCnt);
					for (int k = 0; k < N; k++) {
						if (belongs[k] == deathId) {
							belongs[k] = livingId;
						}
					}
				}
			}
		}
		var res = 0;
		grpCnt.entrySet().stream().max((o1,o2)->Math.max(o1.getValue(),o2.getValue()));

		System.out.println(res);
	}
}