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
  def initialize(n)
    @data = Array.new(n,&:itself)
  end
  def root(i)
    _compact!(i, _root(i))
  end
  def same?(i,j)
    root(i) == root(j)
  end
  def unite!(i,j)
    _compact!(i, root(j))
  end
  def _root(i)
    i = @data[i] until i == @data[i]
    return i
  end
  def _compact!(i,r)
    until @data[i] == r
      tmp = @data[i]
      @data[i] = r
      i = tmp
    end
    return r
  end
end
N, M = gets.split.map(&:to_i)
uf = UnionFind.new(N + 1)
M.times do
  a,b = gets.split.map(&:to_i)
  uf.unite!(a,b)
end
G = (1 .. N).group_by{|i| uf.root(i) }
puts G.transform_values(&:size).values.max