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)

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

@ind_arr = Array.new(n+1){|i| i}
@f_arr = Array.new(n+1, 1)

def find(x)
  return x if x == @ind_arr[x]
  # 途中のindexも更新しておく
  @ind_arr[x] = find(@ind_arr[x])
  return @ind_arr[x]
end

def union(x, y)
  x = find(x)
  y = find(y)
  return if x == y
  @ind_arr[x] = y
  @f_arr[y] += @f_arr[x]
end

m.times do
  a, b = gets.split(" ").map(&:to_i)
  union(a, b)
end

puts @f_arr.max