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)

n, m = gets.strip.split.map(&:to_i)

info = Array.new(n).map { Array.new }

m.times do
  a, b = gets.strip.split.map { |_| _.to_i - 1 }
  info[b] << a
  info[a] << b
  info[a].uniq!
  info[b].uniq!
end

loop do
  infoc = info.dup
  n.times do |i|
    info[i].each do |j|
      t = info[j].dup
      t.delete(i)
      info[i].concat(t).sort!.uniq!
    end
  end
  break if infoc == info
end

ml = info.max_by { |v| v.length }
puts ml.length + 1