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)

from collections import defaultdict

N, M = map(int, input().split())
friend_info = [list(map(int, input().split())) for _ in range(M)]

relation_dict = {}
for person1, person2 in friend_info:
    add_flag = False
    for k, v in relation_dict.items():
        if((person1 in v) or (person2 in v)):
            relation_dict[k].add(person1)
            relation_dict[k].add(person2)
            add_flag = True
            break
    if(add_flag == False):
        new_key = str(len(relation_dict)+1)
        relation_dict[new_key] = {person1, person2}

max_group = max([len(relation) for relation in relation_dict.values()])

print(max_group)