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(:par, :size)

  def initialize(n)
    @par = Array.new(n+1)
    (n+1).times do |i|
      @par[i] = i
    end
    @size = Array.new(n+1,1)
  end

  def root(x)
    while @par[x] != x
      x = @par[@par[x]]
      @par[x] = @par[@par[x]]
    end
    x
  end

  def unite(x,y)
    rx = root(x)
    ry = root(y)
    if rx != ry
      @size[ry] += size(rx)
      @par[rx] = ry
    end
  end

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

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

end
N,M = gets.chomp.split(" ").map(&:to_i)
a,b = M.times.map{gets.split.map(&:to_i)}.transpose
tree = Unionfind.new(N)
M.times do |i|
  tree.unite(a[i],b[i])
end
max = 0
N.times do |i|
  max = [tree.size(i+1),max].max
end
puts max