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: Runtime Error)

N,M=map(int,input().split())
friendlist={i+1:0 for i in range(N)}
sumg=0
for m in range(M):
  A,B=map(int,input().split())
  if friendlist[A]==0 and friendlist[B]==0:
    sumg+=1
    friendlist[A]=sumg
    friendlist[B]=sumg
  elif friendlist[A]==0:
    friendlist[A]=friendlist[B]
  elif friendlist[B]==0:
    friendlist[B]=friendlist[A]
  elif friendlist[A]!=friendlist[B]:
    k=friendlist[B]
    for i in range(1,N+1):
      if friendlist[i]==k:
        friendlist[i]=friendlist[A]
maxlist=[0 for _ in range(sumg+1)]
for i in range(1,N+1):
  maxlist[friendlist[i]]+=1
print(max(maxlist[1:]))