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 Ruby (2.7.1) (judge verdict: Wrong Answer)

(n,m)=gets.split.map(&:to_i)
friend_list = ARGF.map{|s|s.split.map(&:to_i).sort}.sort.uniq

tmp=(0...n).map{[]}
friend_list.each{|a,b|
    tmp[a-1] << b-1
    tmp[b-1] << a-1
}

tmp2=(0...n).map{[]}
done_memo=[false]*n
loop do
    idx = done_memo.index{|x|x==false}
    if idx == nil
        break
    end
    queue=tmp[idx]
    done_memo[idx]=true

    loop do
        if queue.size ==0
            break
        end
        r = queue.shift
        if done_memo[r] == false
            done_memo[r]=true
            queue+=tmp[r]
            tmp2[idx] << r
        end
    end
end
p tmp2

p tmp2.map{|a|a.size+1}.max