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)

class UnionFind
  attr_accessor(:table)

  def initialize(n)
    @table = Array.new(n, -1)
  end

  def root(x)
    st = []
    tb = @table
    while tb[x] >= 0
      st.push(x)
      x = tb[x]
    end
    
    st.each do |y|
      @table[y] = x
    end
    
    return x
  end
  
  def find(x, y)
    return root(x) == root(y)
  end

  def union(x, y)
    r1 = root(x)
    r2 = root(y)
    return if r1 == r2
    
    d1 = @table[r1]
    d2 = @table[r2]
    if d1 <= d2
      @table[r2] = r1
      @table[r1] += d2
    else
      @table[r1] = r2
      @table[r2] += d1
    end
  end
end

n, m = gets.chomp.split.map(&:to_i)
uf = UnionFind.new(n)
m.times do |i|
  a, b = gets.chomp.split.map(&:to_i)
  uf.union(a - 1, b - 1)
end
puts uf.table.min * (- 1)