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 Python (3.8.2) (judge verdict: Runtime Error)

n, m = map(int, input().split())
count = []
A = []
answer = []

for idx in range(m):
    a, b = map(int, input().split())
    if a < b:
        friends = [a, b]
    else:
        friends = [b, a]
    
    if friends not in A:
        A.append(friends)

A = sorted(A)
A_length = len(A)
sub_list = []
for idx in range(A_length):
    a, b = A[idx][0], A[idx][1]
    if idx == 0:
        sub_list += A[idx]
    elif idx == A_length - 1:
        if A[idx - 1][0] == A[idx][0]:
            sub_list += A[idx]
            sub_list = list(set(sub_list))
            answer.append(sub_list)
        else:
            answer.append(sub_list)
            sub_list = []
            sub_list += A[idx]
            answer.append(sub_list)
    else:
        if A[idx - 1][0] == A[idx][0]:
            sub_list += A[idx]
            sub_list = list(set(sub_list))
        else:
            answer.append(sub_list)
            sub_list = []
            sub_list += A[idx]

for ans in answer:
    count.append(len(ans))

print(max(max(count), n - sum(count)))