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}

groups = []
in_group = {}
input.each do |l|
  a, b = l.chomp.split("\s").map{|x| x.to_i}
  if in_group[a].nil? && !in_group[b].nil?
    in_group[a] = in_group[b]
    groups[in_group[b]].push(a)
  elsif !in_group[a].nil? && in_group[b].nil?
    in_group[b] = in_group[a]
    groups[in_group[a]].push(b)
  elsif !in_group[a].nil? && !in_group[b].nil?
    next if in_group[a] == in_group[b]
    groups[in_group[b]].each do |member|
      groups[in_group[a]].push(member) unless groups[in_group[a]].include?(member)
    end
    groups[in_group[b]] = []
    in_group[b] = in_group[a]
    next
  else
    groups.push([a, b])
    in_group[a] = groups.length - 1
    in_group[b] = groups.length - 1
  end
end

cnt = 0
groups.each do |g|
  cnt = g.length if cnt < g.length
end

cnt = 1 if cnt == 0

puts cnt