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 Common Lisp (SBCL 2.0.3) (judge verdict: Accepted)

(let* ((n (read))
       (m (read))
       (stack)
       (c (make-array (list (1+ n)) :initial-element nil))
       (r (make-array (list (1+ n)) :initial-element nil))
       (num (make-array (list (1+ n)) :initial-element 0)))
  (loop :for _ :from 1 :to m
        :for a := (read)
        :for b := (read)
        :do (push a (aref r b))
        :do (push b (aref r a)))
  (defun cnt (x gn)
    (unless (aref c x)
      (setf (aref c x) t)
      (incf (aref num gn))
      (loop :for y :in (aref r x)
            :do (push y stack))))
  (loop :for x :from 1 :to n
        :if (null (aref c x))
        :do (push x stack)
        :do (loop :while stack
                  :for z := (pop stack)
                  :do (cnt z x)))
  (format t "~A~%" (loop :for i :from 1 :to n
                         :maximize (aref num i))))