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: Accepted)

from collections import defaultdict
s = input().split()
n = int(s[0]) 
m = int(s[1])
graph = defaultdict(set)
for i in range(m):
    s = input().split(" ")
    graph[int(s[0])].add(int(s[1]))
    graph[int(s[1])].add(int(s[0]))

visited = [False] * (n+1)

def bfs (node):
    queue = [node]  
    ans = 0
    visited[node] = True
    while queue:
        root = queue.pop()
        ans +=1
        
        for child in graph[root]:
            if not visited[child]:
                queue.append(child)
                visited[child]  =True
    return ans

ans  = 0
for i in range(1, n+1):
    if not visited[i]:
        ans = max(ans, bfs(i))
print( ans)