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)

def connected_components(graph)
	components = []
	visited = [false] * graph.size
	graph.size.times do |i|
		next if visited[i]
		visited[i] = true
		queue = [i]
		component = []
		until queue.empty?
			j = queue.pop
			component << j
			graph[j].each do |k|
				unless visited[k]
					visited[k] = true
					queue << k
				end
			end
		end
		components << component
	end
	components
end

n, m = gets.split.map &:to_i
graph = n.times.map {[]}
m.times do
  a, b = gets.split.map &:to_i
  a -= 1
  b -= 1
  graph[a] << b
  graph[b] << a
end

p connected_components(graph).map(&:size).max