output_description: Print the answers in order, with space in between.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N M K
      A_1 B_1
      \vdots
      A_M B_M
      C_1 D_1
      \vdots
      C_K D_K
attempt: from collections import deque
  
  n, m, k = list(map(int, input().split()))
  
  
  friendship = {}
  for i in range(m):
      a, b = list(map(int, input().split()))
      if a in friendship.keys():
          friendship[a].append(b)
      else:
          friendship[a] = [b]
  
      if b in friendship.keys():
          friendship[b].append(a)
      else:
          friendship[b] = [a]
  
  block = {}
  for j in range(k):
      c, d = list(map(int, input().split()))
      if c in block.keys():
          block[c].append(d)
      else:
          block[c] = [d]
  
      if d in block.keys():
          block[d].append(c)
      else:
          block[d] = [c]
  
  ans = [0] * n
  for i in range(n):
      print(i + 1)
      person = i + 1
      checked_person = set()
      checked_person.add(person)
      if person in friendship.keys():
          candidate = deque(friendship[person])
          while len(candidate) > 0:
              c = deque.popleft(candidate)
              checked_person.add(c)
              if c not in friendship[person]:
                  if person in block.keys():
                      if c not in block[person]:
                          ans[i] += 1
                  else:
                      ans[i] += 1
              if c in friendship.keys():
                  for p in friendship[c]:
                      if p not in checked_person and p not in candidate:
                          candidate.append(p)
      else:
          continue
  
  answer = ""
  for i in range(len(ans)):
      answer += str(ans[i])
      if i != n - 1:
          answer += " "
  
  print(answer)
  
problem_description: Statement
  
  An SNS has N users - User 1, User 2, \cdots, User N.
  
  Between these N users, there are some relationships - M _friendships_ and K
  _blockships_.
  
  For each i = 1, 2, \cdots, M, there is a bidirectional friendship between User
  A_i and User B_i.
  
  For each i = 1, 2, \cdots, K, there is a bidirectional blockship between User
  C_i and User D_i.
  
  We define User a to be a _friend candidate_ for User b when all of the
  following four conditions are satisfied:
  
    * a \neq b.
    * There is not a friendship between User a and User b.
    * There is not a blockship between User a and User b.
    * There exists a sequence c_0, c_1, c_2, \cdots, c_L consisting of integers between 1 and N (inclusive) such that c_0 = a, c_L = b, and there is a friendship between User c_i and c_{i+1} for each i = 0, 1, \cdots, L - 1.
  
  For each user i = 1, 2, ... N, how many friend candidates does it have?
samples: [{"input": "4 4 1\n    2 1\n    1 3\n    3 2\n    3 4\n    4 1", "output": "0 1 0 1\n    \n\nThere is a friendship between User 2 and 3, and between 3 and 4. Also, there\nis no friendship or blockship between User 2 and 4. Thus, User 4 is a friend\ncandidate for User 2.\n\nHowever, neither User 1 or 3 is a friend candidate for User 2, so User 2 has\none friend candidate.\n\n* * *"}, {"input": "5 10 0\n    1 2\n    1 3\n    1 4\n    1 5\n    3 2\n    2 4\n    2 5\n    4 3\n    5 3\n    4 5", "output": "0 0 0 0 0\n    \n\nEveryone is a friend of everyone else and has no friend candidate.\n\n* * *"}, {"input": "10 9 3\n    10 1\n    6 7\n    8 2\n    2 5\n    8 4\n    7 3\n    10 9\n    6 4\n    5 8\n    2 6\n    7 5\n    3 1", "output": "1 3 5 4 3 3 3 3 1 0"}]