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: Wrong Answer)

n,m =map(int, input().split())

list0=[]
list1=[]

listA3=[]
listB3=[]
listA2=[]
listB2=[]

for i in range(m):
  a,b=map(int, input().split())
  c=max(a,b)
  d=min(a,b)
  list0.append((d,c))
  list0.sort()
  list1=set(list0)

for i in list1:
  a2=str(i[0])
  b2=str(i[1])
  listA2.append(a2)
  listB2.append(b2)
  
for i in range(n):
  a3=listA2.count(str(i))
  listA3.append(int(a3))
  b3=listB2.count(str(i))
  listB3.append(int(b3))

ans=0
ans=max(listB3+listA3)

if ans==1:
  print(1)
else:
  ans=ans+1
  print(ans)