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: Accepted)

from collections import*
class UnionFind():
  def __init__(self, n=10**6):
    self.n=n
    self.root=[-1]*(n+1)
    self.rnk=[0]*(n+1)
  def find(self, x):
    if(self.root[x]<0):
      return x
    else:
      self.root[x]=self.find(self.root[x])
      return self.root[x]
  def unite(self, x, y):
    x=self.find(x)
    y=self.find(y)
    if(x==y):return 
    elif(self.rnk[x] > self.rnk[y]):
      self.root[x]+=self.root[y]
      self.root[y]=x
    else:
      self.root[y]+=self.root[x]
      self.root[x]=y
      if(self.rnk[x]==self.rnk[y]):
        self.rnk[y]+=1
  def same(self, x, y):
    return self.find(x)==self.find(y)
  def count(self, x):
    return -self.root[self.find(x)]
n,m=map(int,input().split())
uf=UnionFind(n)
for _ in range(m):
  a,b=map(int,input().split())
  uf.unite(a,b)
a=list(Counter([uf.find(i) for i in range(1,n+1)]).values())
print(max(a))