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)

END { load __FILE__ unless $stdin.eof? }

n, m = gets.split.map!(&:to_i)

nbrs = Array.new(n + 1) { [] }
m.times do
  u, v = gets.split.map!(&:to_i)
  nbrs[u] << v
  nbrs[v] << u
end

max_num = 0
(1..n).each do |i|
  q = nbrs[i]
  next unless q

  group = { i => true }
  while (j = q.pop)
    r = nbrs[j]
    next unless r
    group[j] = true
    nbrs[j] = nil
    q.concat(r)
  end
  num = group.size
  max_num = num if num > max_num

  nbrs[i] = nil
end

puts max_num