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: Accepted)

# frozen_string_literal: true


# union find
class UnionFind
  attr_reader :size

  def initialize(n)
    @parents = Array.new(n) { |i| i }
    @ranks = Array.new(n) { 0 }
    @size = Array.new(n) { 1 }
  end

  def root(x)
    if @parents[x] == x
      x
    else
      @parents[x] = root(@parents[x])
    end
  end

  def unite(x, y)
    root_x = root(x)
    root_y = root(y)
    return if root_x == root_y

    if @ranks[root_x] < @ranks[root_y]
      @parents[root_x] = root_y
      @size[root_y] += @size[root_x]
      @size[root_x] = 0
    else
      @parents[root_y] = root_x
      @size[root_x] += @size[root_y]
      @size[root_y] = 0
      @ranks[root_x] += 1 if @ranks[root_x] == @ranks[root_y]
    end
  end

  def same?(x, y)
    root(x) == root(y)
  end
end

n, m = gets.split.map(&:to_i)
ab = m.times.map { gets.split.map(&:to_i) }

u = UnionFind.new(n)

ab.each do |e|
  u.unite(e[0] - 1, e[1] - 1)
end

puts u.size.max