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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

N, M = map(int, input().split())
 
AB = [list(map(int, input().split())) for _ in range(M)]
 
friend = list()
for a, b in AB:
  a_set = -1
  b_set = -1
  a_f = False
  b_f = False
  for i, f in enumerate(friend):
    if a in f:
      a_set = i
      a_f = True
      if b_f:
        break
    elif b in f:
      b_set = i
      b_f = True
      if a_f:
        break
  if a_set == b_set == -1:
    friend.append({a, b})
  elif a_set == -1:
    friend[b_set].add(a)
  elif b_set == -1:
    friend[a_set].add(b)
  else:
    friend[a_set] = friend[a_set] | friend[b_set]
    friend.pop(b_set)

ans = 0
for f in friend:
  ans = max(len(f), ans)
print(ans)