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

def dfs(node):
    if node in visited:
        return False
    visited.add(node)
    for neighbor in adj_list[node]:
        dfs(neighbor)
    return True

N, M = map(int, input().split())

A, B = [], []

adj_list = defaultdict(set)
visited = set([])

count = 1

for i in range(M):
    a, b = map(int, input().split())
    adj_list[a].add(b)
    adj_list[b].add(a)
    visited.add(a)
    visited.add(b)

#for node in range(1, N+1):
#    if dfs(node):
#        count += 1

print(max(map(lambda x: len(x)+1, adj_list.values())) if adj_list else 1)