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
arr=(1..n).to_a
ab=m.times.map { gets.split.map &:to_i }
friend_list = Hash.new { |h,k| h[k] = Array.new } 
#p ab
m.times do |i|
  #p ab[i]
  #puts "ab[0]: #{ab[i][0]}, ab[1]: #{ab[i][1]}" 
  friend_list[ab[i][0]] << ab[i][1] unless friend_list[ab[i][0]].include? ab[i][1]
  friend_list[ab[i][1]] << ab[i][0] unless friend_list[ab[i][1]].include? ab[i][0]
end
#p friend_list
=begin
past = []
friend_list.each do |k,v|
  friend_list[k] = arr - v  - [k] - past
  past += friend_list[k]
end
p friend_list
=end
p res =  friend_list.sort_by { |_k,v| v.size }.to_h.values.last.size + 1