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 Node
    attr_accessor :val
    def initialize(v, p)
        @val = v
        @parent = p
    end
    def root
        @parent == nil ? self : @parent.root
    end
end


n,m= gets.split(" ").map(&:to_i)
f = []
(1..m).each do
    a,b = gets.chomp.split(" ").map(&:to_i)
    f << [a-1,b-1].sort
end

class UnionFind
    # n:個数
    def initialize(n)
        @count = Array.new(n){|i| 1}
        @parent = Array.new(n){|i| i}
    end

    def root(i)
        if @parent[i] == i
            return i
        else
            r = self.root(@parent[i])
            if @count[i] > 0
                @count[r] += @count[i]
                @count[i] = 0
            end
            @parent[i] = r
            return @parent[i]
        end
    end

    def unite(a,b)
        a = self.root(a)
        b = self.root(b)
        if a != b
            @parent[b] = a
            @count[a] += @count[b]
            @count[b]=0
        end 
    end

    def show
        puts "parents"
        puts @parent.join(",")
        puts "count"
        puts @count.join(",")
    end 

    def max
        return @count.max
    end
end

un = UnionFind.new(n)
f.each do |p|
    un.unite(p[0], p[1])
end
puts un.max
#puts un.show