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.InputStreamReader;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		int n = in.nextInt();
		int m = in.nextInt();
		if (m == 0) {
			System.out.println(1);

		} else {
			int[] a = new int[m];
			int[] b = new int[m];
			for (int i = 0; i < m; i++) {
				a[i] = in.nextInt();
				b[i] = in.nextInt();
			}
			int[] f = new int[n + 1];
			int[] c = new int[n + 1];
			setFrend(f, c, a, b);
			int ans = 1;
			for (int i = 1; i <= n; i++)
				if (c[i] > ans)
					ans = c[i];
			System.out.println(ans);
		}
	}

	private static void setFrend(int[] f, int[] c, int[] a, int[] b) {
		for (int i = 0; i < a.length; i++)
			setPair(f, c, a[i], b[i]);
	}

	private static void setPair(int[] f, int[] c, int a, int b) {
		int at = getTop(f, c, a);
		int bt = getTop(f, c, b);
		if (at == bt)
			return;
		c[at] += c[bt];
		c[bt] = 0;
		f[bt] = at;
	}

	private static int getTop(int[] f, int[] c, int x) {
		if (f[x] == x)
			return x;
		if (f[x] == 0) {
			c[x] = 1;
			f[x] = x;
			return x;
		}
		int t = getTop(f, c, f[x]);
		f[x] = t;
		return t;
	}
}