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)
union_find = [*0..n]
rank = Array.new(n + 1) { 0 }

def union_find.root(idx)
  self[idx] == idx ? idx : (self[idx] = root(self[idx]))
end

m.times do
  a, b = gets.split.map(&:to_i)
  a = union_find.root(a)
  b = union_find.root(b)

  next if a == b

  if rank[a] > rank[b]
    union_find[b] = a
  else
    union_find[a] = b
    rank[b] += 1 if rank[a] == rank[b]
  end
end

h = Hash.new(0)
union_find.each { |e| h[union_find.root(e)] += 1 }

puts h.values.max