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.split.map(&:to_i)
relations = []
m.times do
  relations << gets.split.map(&:to_i)
end

friend_groups = Array.new(1) { Array.new(n - 1) { false } }
friend_groups[0][0] = true

relations.each do |a, b|
  friend_groups.each do |group|
    if group[a-1] || group[b-1]
      group[a-1] = true
      group[b-1] = true
    else
      new_group = Array.new(n - 1) { false }
      new_group[a-1] = true
      new_group[b-1] = true
      friend_groups << new_group
    end
  end
end

puts friend_groups.map { |group| group.count(true) }.max