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)

N,M=map(int,input().split())
A,B=[],[]

l=[-1]*N

def getID(l,idx):
  if idx==-1:
    return -1
  while True:
    if idx==l[idx]:break
    idx=l[idx]
  return idx

d={}
for i in range(M):
  s=list(map(int,input().split()))
  s[0]-=1
  s[1]-=1
  if l[s[0]]==-1 and l[s[1]]==-1:
    l[s[0]]=s[0]
    l[s[1]]=s[0]
    d[s[0]]=2
  elif l[s[0]]==-1 or l[s[1]]==-1:
    for j in range(2):
      if l[s[j]]==-1:
        id=getID(l,s[(j+1)%2])
        l[s[j]]=id
        d[id]+=1
  else:
    r1=getID(l,s[0])
    r2=getID(l,s[1])
    if r1==r2:continue
    l[r1]=r2
    d[r2]+=d[r1]
    d[r1]=0
ans=1
for i in d:
  ans=max(ans,d[i])
print(ans)