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: Accepted)

import collections

class UnionFind():
  def __init__(self, n):
    self.n = n
    self.parent = [i for i in range(self.n)]
    self.rank = [0] * self.n
  
  def find_root(self, x):
    if self.parent[x] == x:
      return x
    else:
      self.parent[x] = self.find_root(self.parent[x])
      return self.parent[x]
  
  def unite(self, x, y):
    rx = self.find_root(x)
    ry = self.find_root(y)
    if rx == ry:
      return
    if self.rank[rx] < self.rank[ry]:
      self.parent[rx] = ry
    else:
      self.parent[ry] = rx
      if self.rank[rx] == self.rank[ry]:
        self.rank[rx] += 1
  
  def is_same(self, x, y):
    return self.find_root(x) == self.find_root(y)

N, M = map(int, input().split())
uf = UnionFind(N)
for _ in range(M):
  A, B = map(lambda x: int(x)-1, input().split())
  uf.unite(A, B)

for i in range(N):
  uf.find_root(i)
c = collections.Counter(uf.parent)
print(c.most_common()[0][1])