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: Runtime Error)

N, M = gets.split.map(&:to_i)
ABs = M.times.map { gets.split.map(&:to_i) }

# $stdout = File.open(File::NULL, "w")

def main
  map = {}
  max = 0
  ABs.each do |ab|
    a, b = ab
    am = map[a]
    bm = map[b]
    newm = if am.nil? && bm.nil?
        { a => true, b => true }
      elsif am.nil? && bm
        bm[a] = true
        bm
      elsif am && bm.nil?
        am[b] = true
        am
      else
        next if am == bm
        am + bm
      end
    max = [max, newm.length].max
    map[a] = newm
    map[b] = newm
  end
  return max
end

STDOUT.puts main

# ACx17 WAx11 TLEx1