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: Time Limit Exceeded)

input = readlines()
n, m = input.shift.chomp.split("\s").map{|x| x.to_i}

list = {}
input.each do |l|
  a, b = l.chomp.split("\s").map{|x| x.to_i}
  list[a] = [] if list[a].nil?
  list[b] = [] if list[b].nil?
  list[a].push(b) unless list[a].include?(b)
  list[b].push(a) unless list[b].include?(a)
end

(1..n).each do |h1|
  list[h1] = [] if list[h1].nil?
  (1..n).each do |h2|
    list[h2] = [] if list[h2].nil?
    next if h1 == h2
    if (list[h1] & list[h2]).length > 0
      list[h1].push(h2) unless list[h1].include?(h2)
      list[h2].push(h1) unless list[h2].include?(h1)
    end
  end
end

groups = []
ng_group = {}
list.each do |k, nglist|
  if groups.length == 0
    groups.push([k])
    nglist.each do |h|
      ng_group[h] = [] if ng_group[h].nil?
      ng_group[h].push(0)
    end
    next
  end
  in_group = nil
  ng_group[k] = [] if ng_group[k].nil?
  groups.each_with_index do |g, i|
    next if ng_group[k].include?(i)
    groups[i].push(k)
    in_group = i
    break unless in_group.nil?
  end
  if in_group.nil?
    groups.push([k])
    in_group = groups.length - 1
  end
  nglist.each do |h|
    ng_group[h] = [] if ng_group[h].nil?
    ng_group[h].push(in_group)
  end
end

puts groups.length