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
s={}
par=[*0...n]
fr=->x{
  u=par[x]
  return u==x ? x : par[x]=fr[u]
}
fu=->x,y{
  x=fr[x]
  y=fr[y]
  par[y]=x if x!=y
}
m.times{
  a,b=gets.split.map{|e| e.to_i-1 }.minmax
  r=s[a]||={}
  next if r[b]
  r[b]=true
  fu[a,b]
}
c={}
n.times{|e|
  r=fr[e]
  c[r]=1+(c[r]||0)
}
puts c.values.max