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.chomp.split(' ').map(&:to_i)
a = Array.new(m)
m.times { |i| a[i] = gets.chomp.split(' ').map { |e| e.to_i - 1 } }

adjs = Array.new(n){ [] }
a.each do |ary|
    s, t = ary
    adjs[s].push(t)
    adjs[t].push(s)
end
adjs.each { |e| e.uniq! }

fGBelongings = Array.new(n)
fGMembers = []
fGSizes = []
fGNum = 0
visiteds = Array.new(n)
n.times do |i|
    next if visiteds[i]
    stack = [i]
    fGMembers[fGNum] = []
    until stack.empty? do
        h = stack.pop
        visiteds[h] = true
        unless fGBelongings[h] then
            fGBelongings[h] = fGNum
            fGMembers[fGNum].push(h)
        end
        adjs[h].each do |f|
            stack.push(f) unless visiteds[f]
        end
    end
    fGSizes[fGNum] = fGMembers[fGNum].size
    fGNum += 1
end
puts fGSizes.max