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)

require "set"
n,m = gets.chomp.split(" ").map(&:to_i)

set = Set.new

hash = {}
m.times do
  a,b = gets.chomp.split(" ").map(&:to_i)
  a,b = b,a if b < a
  hash[a] ||= Set.new
  hash[a] << b
  hash[b] ||= Set.new
  hash[b] << a
end

result = 1
(1..n).each do |i|
  next if set.include?(i) || hash[i].nil?
  arr = [i]
  count = 0
  while arr.length > 0 do
    a = arr.shift
    next if set.include?(a)
    set << a
    count += 1
    arr += hash[a].to_a
  end
  result = count if count > result
end

puts result