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)

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

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

def findChildren(edges, roots, current)
  res = []
  roots.each do |r|
    next if edges[r].nil?
    edges[r].each do |c|
      next if current.include?(c)
      res.push(c) unless res.include?(c)
    end
  end

  res
end
max = 0
edges.each do |k, list|
  members = [k]
  list.each do |l|
    members.push(l)
  end
  while(list.length > 0)
    list = findChildren(edges, list, members)
    list.each do |c|
      members.push(c) unless members.include?(c)
    end
  end
  max = members.length if members.length > max
end

puts max == 0 ? 1 : max