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: Time Limit Exceeded)

import sys
sys.setrecursionlimit(1000000)
n,m = map(int,input().split())
graph = [[] for _ in range(n+1)]
for _ in range(m):
    a,b = map(int,input().split())
    graph[a].append(b)
    graph[b].append(a)
ans = 1
def dfs(node):
    todo = set()
    for nextnode in graph[node]:
        if tf[nextnode] == 0 :
            tf[nextnode] = 1
            seen.add(nextnode)
            todo.add(nextnode)
    for nextnode in todo:
        dfs(nextnode)
done = set()
for i in range(1,n+1):
    if i in done:
        continue
    else:
        done.add(i)
        tf = [0 for _ in range(n+1)]
        seen = set()
        dfs(i)
        ans = max(ans,sum(tf))
        done |= seen
print(ans)