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: Runtime Error)


class Node
    attr_accessor :val
    def initialize(v, p)
        @val = v
        @parent = p
    end
    def root
        @parent == nil ? self : @parent.root
    end
end


n,m= gets.split(" ").map(&:to_i)
f = Hash.new{|h,k| h[k]=[]}
(1..m).each do
    a,b = gets.chomp.split(" ").map(&:to_i)
    f[a] << b unless f[a].include?(b)
    #f[b] << a unless f[b].include?(a)
end

class Friends
    def initialize(f,n)
        @table=f
        @checked = Array.new(n, nil)
        @group = Hash.new{|h,k| h[k]=[]}
        @table.each do |k,v|
            self.add(k)
        end
    end

    def add(key, id=key)
        return false if @checked[key] != nil

        id = id || key
        @checked[key] = id
        @group[id] << key
        @table[key].each do |f|
            self.add(f,id)
        end
        return true        
    end

    def show
        @group.each do |key, ids|
            puts "["+key.to_s+"]"
            puts ids.join(",")
        end
    end 

    def max
        cnt=0
        @group.each do |k,v|
            l = v.length
            cnt = l if l > cnt 
        end
        return cnt
    end
end

fr= Friends.new(f,n)
puts fr.max