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)

n,m = gets.split.map &:to_i
$abs = Array.new(n).map{ Hash.new }

m.times{
	a,b = gets.split.map &:to_i
	$abs[a-1][b-1] = true
	$abs[b-1][a-1] = true
}

$walked = Array.new(m, false)

def walk(start)
	next_n = [start]
	$walked[start] = true
	size = 1
	
	while next_n.size > 0
		temp = []
		next_n.each{ |node|
			$abs[node].keys.each{ |k|
				if ! $walked[k] then
					temp << k
					$walked[k] = true
					size += 1
				end
			}
		}
		next_n = temp
	end

	#print size
	size
end

max = 1

$abs.each_with_index{ |nodes, i|
	if nodes.size > 0 && $walked[i] == false then
		size = walk(i)
		if max < size then
			max = size
		end
	end
}

puts max