output_description: Print the answer.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N M
      A_1 B_1
      \vdots
      A_M B_M
attempt: class Graph:
      def __init__(self, n, m, side):
          self.side = side
          self.neighbor = [set() for i in range(n + 1)]
          self.visited = [False] * (n + 1)
          self.product()
          self.group = [0] * (n + 1)
  
      def product(self):
          for a, b in self.side:
              self.neighbor[a].add(b)
              self.neighbor[b].add(a)
  
      def dfs(self, x, a):
          self.group[a] += 1
          self.visited[x] = True
          for i in self.neighbor[x]:
              if self.visited[i]:
                  continue
              self.dfs(i, a)
  
      def all_dfs(self):
          a = 0
          for i in range(1, n + 1):
              if not self.visited[i]:
                  self.dfs(i, a)
                  a += 1
  
  
  n, m = map(int, input().split())
  side = list(tuple(map(int, input().split())) for i in range(m))
  graph1 = Graph(n, m, side)
  graph1.all_dfs()
  print(max(graph1.group))
  
problem_description: 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?
samples: [{"input": "5 3\n    1 2\n    3 4\n    5 1", "output": "3\n    \n\nDividing them into three groups such as \\\\{1,3\\\\}, \\\\{2,4\\\\}, and \\\\{5\\\\}\nachieves the goal.\n\n* * *"}, {"input": "4 10\n    1 2\n    2 1\n    1 2\n    2 1\n    1 2\n    1 3\n    1 4\n    2 3\n    2 4\n    3 4", "output": "4\n    \n\n* * *"}, {"input": "10 4\n    3 1\n    4 1\n    5 9\n    2 6", "output": "3"}]