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())
ab = [[]]*m
for i in range(m):
  ab[i] = list(map(int,input().split()))
  ab[i].sort()
if m<1:
  print(1)
else:  
  ab.sort()
  frgrps = {0:{ab[0][0],ab[0][1]}}
  grpno = 0
  for i in range(m):
    if ab[i][0] in frgrps[grpno] or ab[i][1] in frgrps[grpno]:
      frgrps[grpno].update(ab[i])
    else:
      grpno += 1
      frgrps[grpno] = {ab[i][0],ab[i][1]}
longestset = 0
for x in frgrps:
  longestset = max(longestset,len(frgrps[x]))
print(longestset)