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

public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String str[] = br.readLine().split(" ");

		int n = Integer.parseInt(str[0]);
		int m = Integer.parseInt(str[1]);

		UnionFind uft = new UnionFind(n);

		for (int i = 0; i < m; i++) {
			str = br.readLine().split(" ");

			int a = Integer.parseInt(str[0]);
			int b = Integer.parseInt(str[1]);

			uft.unite(a - 1, b - 1);
		}

		long max = 0;
		for (int i = 0; i < n; i++) {
			max = Math.max(max, uft.size(i));
		}

		System.out.println(max);
	}
}

class UnionFind {

	// 自身が親であれば、その集合に属する頂点数に-1を掛けたもの
	// そうでなければ親のid
	int r[];

	public UnionFind(int n) {
		r = new int[n];
		Arrays.fill(r, -1);
	}

	int root(int x) {
		if (r[x] < 0)
			return x;
		return r[x] = root(r[x]);
	}

	boolean unite(int x, int y) {
		x = root(x);
		y = root(y);
		if (x == y)
			return false;
		if (r[x] > r[y]) {
			int tmp = x;
			x = y;
			y = tmp;
		}
		r[x] += r[y];
		r[y] = x;
		return true;
	}

	int size(int x) {
		return -r[root(x)];
	}
}