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)

pairs = Array.new(n){Array.new(0)}

num = 0

for i in 0...m
  tmp = gets.split.map(&:to_i)
  pairs[tmp.min-1].push(tmp.max-1)
end

for i in 0...n
  pairs[i] = pairs[i].uniq.sort
end

group = Array.new(n,nil)
group[0] = 0

for i in 0...n
  if group[i]==nil
    kouho = i
    for j in pairs[i]
      if group[j]!=nil
        kouho = [kouho,group[j]].min
      end
    end
    group[i] = kouho
  end
  
  for j in pairs[i]
    if group[j]==nil
      group[j] = group[i]
    end
  end
end

ans = 0

group = group.sort
tmp = 1
for i in 1...n
  if group[i]==group[i-1]
    tmp += 1
  else
    ans = [ans,tmp].max
    tmp = 1
  end
end

ans = [ans,tmp].max

puts ans