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: Time Limit Exceeded)

N,M=gets.split.map(&:to_i)
l=Array.new(N){[]}
M.times{
    a,b=gets.split.map{|i|i.to_i-1}
    l[a] << b
    l[b] << a
}
l.map!{|i|i.uniq}
warn l.inspect
comp=Array.new(N)
closed=Array.new(N){false}
s=[]
N.times{|i|
    next if closed[i]
    s.push(i)
    until s.empty?
        v=s.pop
        closed[v]=true
        comp[v]=i
        l[v].each{|n|
            next if closed[n]
            s.push(n)
        }
    end
}
count=Hash.new{0}
max=0
comp.each{|i|
    count[i]+=1
    if max<count[i]
        max=count[i]
    end
}
puts max