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)
l=[]
n.times do
  l << []
end
m.times do
  a,b=gets.split.map(&:to_i)
  l[a-1] << b-1
  l[b-1] << a-1
end
v=[0]*n
x=0
n.times do |i|
  if v[i]==0
    v[i]=1
    s=[i]
    c=1
    while s!=[]
      z=[]
      s.each do |ss|
        l[ss].each do |lss|
          if v[lss]==0
            v[lss]=1
            c+=1
            z << lss
          end
        end
      end
      s=z.dup
    end
    x=c if x<c
  end
end
puts x