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: Time Limit Exceeded)

from collections import deque

n,m = map(int,input().split())
ab = [list(map(int,input().split())) for _ in range(m)]

g = {i:[] for i in range(n+1)}

for a,b in ab:
    g[a].append(b)
    g[b].append(a)
    
seen = {i:0 for i in range(n+1)}
seen[0]=1
seen[1]=1
q = deque()
q.append(1)
grpsize=1

while 0 in seen.values():
    if len(q)>0:
        now = q.popleft()
        nextp = g[now]
        for i in nextp:
            if seen[i]==0:
                q.append(i)
                grpsize+=1
                seen[i]=grpsize
        
    else:
        for i in range(1,n+1):
            if seen[i]==0:
                grpsize=1
                seen[i]=grpsize
                q.append(i)
                break
                
print(max(seen.values()))