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: Time Limit Exceeded)

class UnionFind
  def initialize(n)
    @n = n
    @par = Array.new(@n).map.with_index{|x, i| i}
    @chl = Array.new(@n).map{[]}
  end
  def union(a, b)
    pa, pb = [@par[a], @par[b]]
    if pa < pb then
      absorb(a, b)
    elsif pa > pb then
      absorb(b, a)
    end
  end
  def max_size
    @chl.compact.map{|x| x.size}.max
  end
  
  private
  
  def absorb(pa, ch)
    par = @par[pa]
    chl = @par[ch]
    @chl[chl].each{|c|
      @chl[par] << c
      @par[c] = par
    }
    @chl[chl] = nil
    @chl[par] << chl
    @par[chl] = par
  end
end

N, M = gets.chomp.split(" ").map(&:to_i)
R = (0...M).map{gets.chomp.split(" ").map(&:to_i)}

uf = UnionFind.new(N+1)
R.each{|a, b| uf.union(a, b)}

puts uf.max_size+1