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: Accepted)

lines = []
while line = gets
    lines << line.chomp.split(' ').map(&:to_i)
end

N = lines[0][0]
M = lines[0][1]

access = []
for i in 1..N do
  access[i-1] = []
end

for i in 1..M do
  from = lines[i][0]
  to = lines[i][1]
  access[from-1].push(to)
  access[to-1].push(from)
end

visited = []
for i in 1..N do
  visited[i-1] = false
end

num_member = 0
max_num_member = 0
for i in 1..N do
  if visited[i-1] == false then
    num_member = 0
    targets = [i]
    while targets != [] do
      num_member += 1
      j = targets.pop
      visited[j-1] = true
      access[j-1].each do |k|
        if visited[k-1] == false then
          targets.push(k)
          visited[k-1] = true
        end
      end
    end
    if max_num_member < num_member then
      max_num_member = num_member
    end
  end
end
    
puts max_num_member