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)

# Verified: https://atcoder.jp/contests/atc001/submissions/16154602
class UnionFind
	Node = Struct.new(:p, :h)
	def initialize(size)
		@p, @r = size.times.to_a, [0]*size
	end
	def unite(i, j)
		k, l = parent(i), parent(j); if @r[k] < @r[l]; @p[k] = l; @r[l] += 1 if @r[k] == @r[l]; else; @p[l] = k; end
	end
	def united?(i, j)
		parent(i) == parent(j)
	end
	def parent(i)
		j = i; until i == @p[i]; j, i = i, @p[j] = @p[i]; end; i
	end
end

n, m = gets.split.map &:to_i
uf = UnionFind.new n
m.times do
	a, b = gets.split.map &:to_i
	uf.unite a-1, b-1
end

parents = [0] * n
n.times do |i|
	parents[uf.parent i] += 1
end

p parents.max