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 :rank, :parent, :size
  def initialize(s)
    @rank = Array.new(s,0)
    @parent = Array.new(s,&:itself)
    @size = Array.new(s,1)
  end

  def union(x,y)
    x_parent = get_parent(x)
    y_parent = get_parent(y)
    return if x_parent == y_parent

    if rank[x_parent] > rank[y_parent]
      parent[y_parent] = x_parent
      size[x_parent] += size[y_parent]
    else
      parent[x_parent] = y_parent
      size[y_parent] += size[x_parent]
      rank[y_parent] += 1 if rank[x_parent]==rank[y_parent]
    end
  end

  def get_parent(id)
    parent[id] == id ? id : parent[id] = get_parent(parent[id])
  end
end

n,m=gets.split.map(&:to_i)
u = UnionFind.new(n)
m.times do
  a,b = gets.split.map(&:to_i)
  u.union(a-1,b-1)
end
p u.size.max