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: Wrong Answer)

N, M = gets.split.map(&:to_i)

groups = Hash.new()
belongs = [nil]*(N + 1)

group = 0
M.times {
  a, b = gets.split.map(&:to_i)
  a_belong, b_belong = [belongs[a], belongs[b]]

  if a_belong != nil and b_belong != nil and a_belong != b_belong then
    groups[a_belong] += groups[b_belong]
    groups[b_belong].each { |i|
      belongs[i] = a_belong
    }
  else
    belong = belongs[a] != nil ? belongs[a] : belongs[b] != nil ? belongs[b] : group

    belongs[a] = belong
    belongs[b] = belong

    if groups[belong] == nil then
      groups[belong] = [a, b]
    else
      groups[belong] += [a, b]
    end
  end

  group += 1
}

p groups.values.map{|list| list.uniq.size }.max