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
  def initialize(n)
    @size = Array.new(n, 1)
    @rank = Array.new(n, 0)
    @parent = []

    (0..n).each do |i|
      @parent[i] = i
    end
  end

  def find(x)
    if @parent[x] == x
      x
    else
      @parent[x] = find(@parent[x])
    end
  end

  def unite(x, y)
    x = find(x)
    y = find(y)
    return if x == y

    if @rank[x] < @rank[y]
      @parent[x] = y
      @size[y] += @size[x]
    else
      @parent[y] = x
      @size[x] += @size[y]

      @rank[x] += 1 if @rank[x] == @rank[y]
    end
  end

  def same?(x, y)
    find(x) == find(y)
  end

  def size(x)
    @size[find(x)]
  end
end

N, M = gets.split.map(&:to_i)
Q = M.times.map { gets.split.map(&:to_i) }

uf = UnionFind.new(N + 1)

Q.each do |a, b|
  uf.unite(a, b)
end

puts (1..N).map { |x| uf.size(x) }.max