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 = gets.chomp.split(' ').map(&:to_i)
friends_visited = Array.new(n){false}
groups = {}
group_ref = {}
group_count = 0
m.times do
    a,b = gets.chomp.split(' ').map(&:to_i)
    if friends_visited[a-1] && friends_visited[b-1] && (groups[a-1] != groups[b-1])
        group_ref[groups[a-1]] = group_ref[groups[b-1]]
    elsif !friends_visited[a-1] && friends_visited[b-1]
        friends_visited[a-1] = true
        groups[a-1] = groups[b-1]
    elsif friends_visited[a-1] && !friends_visited[b-1]
        friends_visited[b-1] = true
        groups[b-1] = groups[a-1]
    elsif !friends_visited[a-1] && !friends_visited[b-1]
        friends_visited[a-1] = true
        friends_visited[b-1] = true
        groups[a-1] = group_count
        groups[b-1] = group_count
        group_ref[group_count] = group_count
        group_count += 1
    end
end

count = Array.new(group_count){0}
n.times do |i|
    if friends_visited[i]
        count[group_ref[groups[i]]] += 1
    end
end

puts count.max