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: Time Limit Exceeded)

N, M = map(int,input().split())
pairs = []
for i in [0]*M:
    inp = sorted(list(map(int,input().split())))
    if inp not in pairs:
        pairs.append(inp)
def solver(N,M,pairs):
    group_list = []
    max_len = 1
    for i in pairs:
        pair = sorted(i)
        for j in range(len(group_list)):
            if pair[1] in group_list[j]:
                break
            elif pair[0] in group_list[j]:
                group_list[j].append(pair[1])
                if max_len < len(group_list[j]):
                    max_len = len(group_list[j])
                break
        else:
            if max_len < 2:
                max_len = 2
            group_list.append(pair)
    return max_len
print(solver(N,M,pairs))