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,*P = $<.read.split.map &:to_i
A = [[]]*(N+1)
m = 0
M.times {|i|
  r,a = [],P[i+i,2]
  x,y = a.map {|i;j| (r<< i; i=j) while (j=A[i])[0]; r<< i; i}.sort
  g,h = [A[x],A[y]].map{|h| h.empty? ? a.to_h{[_1,0]} : h}.sort_by{_1.size}
  h.merge! g  if h.object_id != g.object_id
  r.each {A[_1] = y}
  A[y] = h
  m < h.size && m = h.size
}
p m