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: Runtime Error)

require "set"
n, m = gets.split(" ").map(&:to_i)

arr = []
hash = {}

m.times do
  a, b = gets.split(" ").map(&:to_i)
  if arr[a].nil?
    hash[a] = Set.new([a])
    arr[a] = a
  end
  if arr[b].nil?
    hash[b] = Set.new([b])
    arr[b] = b
  end
  a_ind = arr[a]
  b_ind = arr[b]
  hash[a_ind].merge(hash[b_ind])
  hash.delete(b_ind)
  arr[b] = a_ind
end

max = 0
hash.each do |_k, v|
  l = v.length
  max = l if max < l 
end

puts max