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)

from collections import defaultdict

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

not_friend_graph = defaultdict(set)
for a, b in R:
    not_friend_graph[a].add(b)
    not_friend_graph[b].add(a)

max_n = 0
max_i = 0
for i in range(N):
    if len(not_friend_graph[i]) > max_n:
        max_n = len(not_friend_graph[i])
        max_i = i

def dfs_stack(start, visited=None):
    stack = [start]
    visited = set()

    while stack:
        current_vertex = stack.pop()
        if current_vertex not in visited:
            visited.add(current_vertex)
            for v in not_friend_graph[current_vertex]:
                if v not in visited:
                    stack.append(v)
    return visited

visited = dfs_stack(max_i)
print(visited)
print(len(visited))