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)

m,n=map(int,input().split())
a_list=[i+1 for i in range(m)]
m_list=[]
for i in range(n) :
  a,b = map(int, input().split())
  if a<b :
    m_list.append(str(a)+' '+str(b))
  else :
    m_list.append(str(b) +' '+ str(a))

m_list=list(set(m_list))
n1=len(m_list)
s_list=[]
for i in m_list :
  s_list.append(list(map(int, i.split())))

s_list=sorted(s_list,key=lambda x: x[0])





for s in s_list :
  a_list[s[1]-1]=a_list[s[0]-1]
import collections
c = collections.Counter(a_list)


print(max(list(c.values())))