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 {
    private static Scanner sc = new Scanner(System.in);
    public static void main(String[] args) throws Exception {
    int N = sc.nextInt();
    int M = sc.nextInt();       
    UnionFind uf = new UnionFind(N);
    for(int i=0;i<M;i++){
      int A = sc.nextInt()-1;
      int B = sc.nextInt()-1;
      uf.union(A,B);
    }
    System.out.println(uf.maxSize());
}

static class UnionFind {
	
	int[] parent;
	int[] rank;
	
	public UnionFind(int size) {
		this.parent = new int[size];
		this.rank = new int[size];
		for(int i=0;i<size;i++){
          parent[i] = i;
          rank[i] = 1;
        }
	}
	
	public void union(int x, int y) {
		int xRoot = find(x);
		int yRoot = find(y);
		
		//xが属する木の方が大きい場合
		if (rank[xRoot] > rank[yRoot]) { 
			parent[yRoot] = xRoot; //yの親をxに更新
            rank[xRoot]+=rank[yRoot];
		} else if(rank[xRoot] < rank[yRoot]) {
			parent[xRoot] = yRoot;
            rank[yRoot]+=rank[xRoot];
		} else if (xRoot != yRoot){
			parent[yRoot] = xRoot;
			rank[xRoot]+=rank[yRoot]; //同じ高さの木がルートの子として着くから大きさ++;
		}
	}
	
	//引数の属する木のルートを返す
	public int find(int i) {
		if (i != parent[i]) {
			parent[i] = find(parent[i]);
		}
		return parent[i];
	}
	
	//同じ木に属しているか
	public boolean same(int x, int y) {
		return find(x) == find(y);
	}
    
    public int maxSize() {
      int max = 0;
      for(int i=0;i<rank.length;i++){
        //System.out.println(rank[i]);
        max = Math.max(max,rank[i]);
      }
      return max;
	}
  
    public void parList(){
      for(int i=0;i<parent.length;i++){
        System.out.println(i+"の親は"+parent[i]);
      }
      System.out.println();
    }
}
}