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: Time Limit Exceeded)

def uf_find(x):
  if x == p[x]:
    return x
  else:
    p[x] = uf_find(p[x])
    return p[x]

def uf_unite(x,y):
  x = uf_find(x)
  y = uf_find(y)
  if x != y:
    if r[y]>r[x]:
      x,y = y,x
    if r[x]==r[y]:
      r[x]+= 1
    p[y] = x
    #print(r)


N, M = map(int, input().split())
p = list(range(N))
r = [0]*N

for i in range(M):
  A, B = [int(j)-1 for j in input().split()]
  uf_unite(A,B)
  #print(p)
  
#for i in range(N):
  #p[i] = uf_find(i)

q = [0]*N
for i in range(N):
  q[i] = p.count(i)
  
print(max(q))