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: Wrong Answer)

class UnionFind
  def initialize(n)
    @par = (0..n).to_a
    @rank = Array.new(n + 1, 0)
  end
  def root(x)
    @par[x] == x ? x : @par[x] = root(@par[x])
  end
  def par
    @par
  end
  def unite(x, y)
    ux = root(x)
    uy = root(y)
    if @rank[ux] < @rank[uy]
      @par[ux] = uy
    else
      @par[uy] = ux
      @rank[ux] += 1 if @rank[ux] == @rank[uy]
    end
  end  
end

n, m = gets.split.map(&:to_i)
u = UnionFind.new(n)
m.times do
  a, b = gets.split.map(&:to_i)
  u.unite(a, b)
end
puts u.par.count(u.par.group_by{ |e| e }.sort_by{ |k, v| -v.size}.map(&:first).first)