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: Wrong Answer)

n,m = gets.split.map &:to_i
abs = []

m.times{
	a,b = gets.split.map &:to_i
	if a < b then
		abs << [a-1,b-1]
	else
		abs << [b-1,a-1]
	end
}

abs.sort!{ |a,b|
	a[1] <=> b[1]
}

nx = Array.new(n, 0)

(0...n).each{ |i|
	nx[i] = i
}

abs.each{ |a,b|
	if nx[a] < nx[b] then
		nx[b] = nx[a]
	else
		nx[a] = nx[b]
	end
}

print nx.group_by(&:itself).map{ |k,v| v.size}.max