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.Scanner;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        
        int n = sc.nextInt();
        int m = sc.nextInt();

        int groups[] = new int[n];
        int counts[] = new int[n];
        for (int i = 0; i < n; ++i) {
            groups[i] = i;
            counts[i] = 0;
        }

        for (int i = 0; i < m; ++i) {
            int a = sc.nextInt() - 1;
            int b = sc.nextInt() - 1;

            int group_a = groups[a];
            int group_b = groups[b];

            if (group_a != group_b) {
                for (int j = 0; j < n; ++j) {
                    if (groups[j] == group_b) {
                        groups[j] = group_a;
                    }
                }
            }
        }
        
        // Number of members of the largest group
        int max = 0;
        for (int i = 0; i < n; ++i) {
            int g = groups[i];
            counts[g] += 1;
        }
        for (int i = 0; i < n; ++i) {
            max = Math.max(max, counts[i]);
        }

        System.out.println(max);
    }
}