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)

class UnionFind
  attr_accessor :parent

  def initialize(size)
      @parent = Array.new(size, -1)
  end

  def root(x)
      if @parent[x] < 0
          x
      else
          @parent[x] = root(@parent[x])
      end
  end

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

  def merge(x, y)
      x_root = root(x)
      y_root = root(y)

      return false if x_root == y_root

      if @parent[x_root] > @parent[y_root]
          x_root, y_root = y_root, x_root
      end

      @parent[x_root] += @parent[y_root]
      @parent[y_root] = x_root

      return true
  end

  def size(x)
      return -@parent[root(x)]
  end
end

N, M = gets.split.map(&:to_i)
uf = UnionFind.new(N + 1)

m = 0
M.times do |i|
  pair = gets.split.map(&:to_i)
  t = uf.merge(pair[0],pair[1])
end

N.times do |i|
  a = uf.size(i + 1)
  m = a if m < a
end

puts m