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 :arr

  def initialize(size)
    @arr = Array.new(size, -1)
  end

  def unite(a, b)
    pa, pb = [parent(a), parent(b)].sort
    return if pa == pb
    @arr[pb] = pa
  end

  def parent(x)
    return x if @arr[x] == -1
    @arr[x] = parent(@arr[x])
  end

  def same(a, b)
    self.parent(a) == self.parent(b)
  end
end

n, m = gets.chomp.split(' ').map(&:to_i)
uf = UnionFind.new(n + 1)
m.times do
  a, b = gets.chomp.split(' ').map(&:to_i)
  uf.unite(a, b)
end
map = Hash.new(0)
(1..n).each do |i|
  map[uf.parent(i)] += 1
end
puts map.values.max