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: Wrong Answer)

n, m = gets.split.map(&:to_i)
relations = []
m.times do
  relations << gets.split.map(&:to_i).sort
end

# graph = Array.new(n) { [] }
# relations.each do |a, b|
#   graph[a - 1] << b - 1
#   graph[b - 1] << a - 1
# end

# groups = []

# 0.upto(n - 1) do |leader|
#   next if groups[leader]

#   queue = [leader]

#   until queue.empty?
#     current = queue.shift
#     next if groups[current]

#     graph[current].each do |friend|
#       next if groups[friend]

#       queue << friend
#     end

#     groups[current] = leader
#   end
# end

# puts groups.tally.values.max

union_find = (0...n).to_a

def root(union_find, a)
  return a if union_find[a] == a

  union_find[a] = root(union_find, union_find[a])
end

relations.each do |a, b|
  union_find[b - 1] = root(union_find, a - 1)
end

normalized_union_find = (0...n).map { |num| root(union_find, union_find[num])}

pp normalized_union_find.tally.values.max