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)

input = readlines()
n, m = input.shift.chomp.split("\s").map{|x| x.to_i}

tree = Array.new(n, -1)

def getRoot(tree, target)
  while(tree[target] > -1)
    target = tree[target]
  end

  target
end

input.each do |l|
  a, b = l.chomp.split("\s").map{|x| x.to_i - 1}.sort
  a_root = getRoot(tree, a)
  b_root = getRoot(tree, b)
  next if a_root == b_root

  tree[a_root] += tree[b_root]
  tree[b_root] = a_root
end

max = 0
tree.each do |v|
  next if v > -1
  size = v * -1
  max = size if max < size
end

puts max