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: Accepted)

import java.io.*;
import java.util.*;
import java.util.function.*;

final class Solver {
	void solve(Supplier<String> sc) {
		int N = Integer.parseInt(sc.get());
		int M = Integer.parseInt(sc.get());
		var group = new UnionFind(N);
		for (int i = 0; i < M; i++) {
			int A = Integer.parseInt(sc.get())-1;
			int B = Integer.parseInt(sc.get())-1;
			group.union(A, B);
		}
		int max = 0;
		for (int i = 0; i < N; i++) {
			max = Math.max(max, group.size(i));
		}
		System.out.println(max);
	}
}

class UnionFind {
	int[] parent;

	UnionFind(int size) {
		parent = new int[size];
		Arrays.fill(parent, -1);
	}

	int find(int i) {
		if (parent[i] < 0) return i;
		return parent[i] = find(parent[i]);
	}

	int size(int i) {
		return -parent[find(i)];
	}

	boolean same(int i, int j) {
		return find(i) == find(j);
	}

	boolean union(int i, int j) {
		i = find(i);
		j = find(j);
		if (i == j) return false;
		if (size(i) < size(j)) {
			int temp = i;
			i = j;
			j = temp;
		}

		parent[i] += parent[j];
		parent[j] = i;
		return true;
	}
}

class Main {
	public static void main(String... args) {
		System.setOut(new PrintStream(new BufferedOutputStream(System.out)));
		var reader = new BufferedReader(new InputStreamReader(System.in));
		new Solver().solve(new Supplier<>() {
			StringTokenizer line;

			public String get() {
				while (line == null || !line.hasMoreTokens()) try {
					line = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new UncheckedIOException(e);
				}
				return line.nextToken();
			}
		});
		System.out.flush();
	}
}