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,*P = $<.read.split.map &:to_i
C = [m=1]*(N+1)
A = C.map {[]}
(0...M).map {|i| a=P[i+=i]; b=P[i+1]; a<b ? [a,b] : [b,a]}.sort.each{|a,b| A[a] << b}
1.upto(N) {|i|
  a = A[j=i]
  a.each {k = A[_1][-1]||_1; j<k && j=k}
  next  if j == i
  a.each {b=A[_1]; b << j  if b[-1]!=j}
  m < C[j]+=C[i] and m = C[j]
}
p m