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.util.*;

public class Main {
	public static void main (String[] args) {
    	Scanner sc = new Scanner(System.in);
    	int n = sc.nextInt();
    	int m = sc.nextInt();
    	UnionFindTree uft = new UnionFindTree(n);
    	for (int i = 0; i < m; i++) {
    	    uft.unite(sc.nextInt() - 1, sc.nextInt() - 1);
    	}
    	System.out.println(uft.getMax());
    }
    
    static class UnionFindTree {
        int[] parents;
        int[] counts;
        
        public UnionFindTree(int size) {
            parents = new int[size];
            counts = new int[size];
            for (int i = 0; i < size; i++) {
                parents[i] = i;
                counts[i] = 1;
            }
        }
        
        public int find(int x) {
            if (parents[x] == x) {
                return x;
            } else {
                return parents[x] = find(parents[x]);
            }
        }
        
        public boolean smae(int x, int y) {
            return find(x) == find(y);
        }
        
        public void unite(int x, int y) {
            int xx = find(x);
            int yy = find(y);
            if (xx == yy) {
                return;
            }
            parents[xx] = yy;
            counts[yy] += counts[xx];
        }
        
        public int getMax() {
            int max = 0;
            for (int x : counts) {
                max = Math.max(max, x);
            }
            return max;
        }
    }
}