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: Time Limit Exceeded)

n, m = gets.split.map(&:to_i)
gidx = []
g = []
mg_idx = 0
links = m.times.map { gets.split.map(&:to_i).sort }.uniq
if links.empty?
  puts '1'
  exit 0
end
links.each do |a, b|
  if gidx[a] && !gidx[b]
    gidx[b] = gidx[a]
    g[gidx[a]] << b
  elsif !gidx[a] && gidx[b]
    gidx[a] = gidx[b]
    g[gidx[b]] << a
  elsif gidx[a] && gidx[b]
    next if gidx[a] == gidx[b]


    tmp = g[gidx[a]].dup
    g[gidx[a]] = []
    tmp.each do |ggidx|
      gidx[ggidx] = gidx[b]
    end
    g[gidx[b]] = [*g[gidx[b]], *tmp]
  else
    g[mg_idx] = [a, b]
    gidx[a] = mg_idx
    gidx[b] = mg_idx
    mg_idx += 1
  end
end
puts g.map{_1.count}.max