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 Julia (1.4.0) (judge verdict: Wrong Answer)

parseInt(x) = parse(Int64, x)
parseMap(x::Array{SubString{String},1}) = map(parseInt, x)
function main()
  N,M = parseMap(split(readline()))
  set = Set()
  A = zeros(Int64,M); B = zeros(Int64,M)
  for i in 1:M
    A[i],B[i] = parseMap(split(readline()))
    push!(set,A[i]=>B[i])
  end
  count=zeros(N)
  #println(set)
  for item in set
    count[item.first] += 1
    count[item.second] += 1
  end
  if Int64(maximum(count)) == N
    println(Int64(maximum(count)))
  else
    println(Int64(maximum(count)+1))
  end
end
main()