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 Counter
N, M = map(int, input().split())
groups = [0] * (N + 1)
group_num_list = [[0]]
group_num = 1
for _ in range(M):
    A, B = map(int, input().split())
    if groups[A] != 0 and groups[B] != 0:
        if groups[A] != groups[B]:
            lower = min(groups[A], groups[B])
            higher = max(groups[A], groups[B])
            for i in group_num_list[higher]:
                groups[i] = lower
            tmp = group_num_list[higher]
            group_num_list[lower].extend(tmp)
            # del group_num_list[higher]
    elif groups[A] != 0:
        groups[B] = groups[A]
        group_num_list[groups[A]].append(B)
    elif groups[B] != 0:
        groups[A] = groups[B]
        group_num_list[groups[B]].append(A)
    else:
        groups[A] = group_num
        groups[B] = group_num
        group_num_list.append([A, B])
        group_num += 1
counter = Counter(groups)
# print(collection)
# print(group_num_list)
if counter.most_common()[0][0] != 0:
    print(counter.most_common()[0][1])
else:
    print(counter.most_common()[1][1])