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)
if m == 0
  puts 1
  exit
end
a,b = m.times.map{gets.split.map(&:to_i)}.transpose

nds = Array.new(n+1).map{Array.new(0)}
for i in 0..(m-1)
  nds[a[i]].push(b[i])
  nds[b[i]].push(a[i])
end
for i in 1..n
  nds[i].push(-1)
end

ckd = Array.new(n+1,0)
lsin = Array.new(n+1,0)
ans = 0
cntn = 1
stid = 1
while cntn == 1
  ls = Array.new(0)
  ls.push(stid)
  lsln = 1
  lsin[stid] = 1
  for i in 0..(lsln-1)
    ckd[ls[i]] = 1
    nds[ls[i]].each do |ch|
      if ch != -1 && lsin[ch] == 0
        ls.push(ch)
        lsln += 1
        lsin[ch] = 1
      end
    end
  end
  ans = [ans,lsln].max
  
  stid += 1
  while stid < n && ckd[stid] == 1
    stid += 1
  end
  if stid >= n
    cntn = 0
  end
end
puts ans