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)

# 各unionのsizeを持つパターン
class UnionFindWithSize
  def initialize(size)
    @rank = Array.new(size) { 0 }
    @parent = Array.new(size) { |id| id }
    @union_size = Array.new(size) { 1 }
    @max_union_size = 0
  end

  def unite(id_x, id_y)
    x_parent = get_parent(id_x)
    y_parent = get_parent(id_y)
    return if x_parent == y_parent

    ary = [@union_size[x_parent], @union_size[y_parent]]
    if @rank[x_parent] > @rank[y_parent]
      @parent[y_parent] = x_parent
      @union_size[x_parent] += ary[1]
      @max_union_size = @union_size[x_parent] if @union_size[x_parent] > @max_union_size
    else
      @parent[x_parent] = y_parent
      @union_size[y_parent] += ary[0]
      @max_union_size = @union_size[y_parent] if @union_size[y_parent] > @max_union_size
      @rank[y_parent] += 1 if @rank[x_parent] == @rank[y_parent]
    end
    ary
  end

  def get_parent(id_x)
    @parent[id_x] == id_x ? id_x : (@parent[id_x] = get_parent(@parent[id_x]))
  end

  def max_us
    @max_union_size
  end
end

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

ufs = UnionFindWithSize.new(N)
M.times do |i|
  a, b = gets.chomp.split.map(&:to_i)
  ufs.unite(a - 1, b - 1)
end

if M == 0
  puts 1
else
  puts ufs.max_us
end