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: Wrong Answer)

def root(x):
  if par[x] == x:
    return x
  par[x] = root(par[x])
  return par[x]

def union(x, y):
  rx = root(x)
  ry = root(y)
  if rx != ry:
    par[rx] = ry
  return 0
    
n, m=map(int, input().split(" "))
par = list(range(n + 1))
for i in range(m):
  a, b = map(int, input().split(" "))
  union(a, b)
#print(par)
maxi = 0
count = 1
#print(par)
countpar = [0 for i in range(n + 1)]
for i in range(1, n + 1):
  print(i)
print(max(countpar))