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)

a = gets.split(' ')
n = a[0].to_i
m = a[1].to_i

$rr = Array.new(n) {|i| i}
def get_root(u) 
  if u == $rr[u]
    return u 
  else
    return $rr[u] = get_root($rr[u])
  end
end

def merge_root(u, v)
  x = get_root(u)
  y = get_root(v)
  # p x, y
  if (x == y) 
    return 0
  end
  $rr[x] = y
end

m.times do
  a = gets.split(' ')
  u = a[0].to_i - 1
  v = a[1].to_i - 1
  merge_root(u, v)
end

$h = Hash.new
n.times do |i|
  $h.has_key?(get_root(i)) ? $h[get_root(i)] += 1 : $h[get_root(i)] = 1
end

ans = 0
$h.each do |k, v|
  if (v > ans) 
    ans = v
  end
end
puts(ans)