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: import queue
  
  q = queue.Queue()
  n, m, k = map(int, input().split())
  friend = [list(map(int, input().split())) for i in range(m)]
  block = [list(map(int, input().split())) for i in range(k)]
  suggestions = []  # 友達候補
  
  
  def Strand(now):  # 繋がり検索
      for o in range(m):
          if friend[o][0] == now:
              if friend[o][1] not in Q:
                  q.put(friend[o][1])
                  Q.append(friend[o][1])
          if friend[o][1] == now:
              if friend[o][0] not in Q:
                  q.put(friend[o][0])
                  Q.append(friend[o][0])
  
  
  for i in range(n):
      now = i  # 現在地
      count = 0
      Q = [i]  # 発見済み
      Strand(i)
  
      F = []  # 人iのフレンドリスト
      for j in range(m):
          if friend[j][0] == i:
              F.append(friend[j][1])
          if friend[j][1] == i:
              F.append(friend[j][0])
      B = []  # 人iのブロックリスト
      for l in range(k):
          if block[l][0] == i:
              B.append(block[l][1])
          if block[l][1] == i:
              B.append(block[l][0])
  
      while not q.empty():
          if i != 0:
              for o in range(m):
                  if friend[o][0] == now:
                      if friend[o][1] not in Q:
                          q.put(friend[o][1])
                          Q.append(friend[o][1])
                  if friend[o][1] == now:
                      if friend[o][0] not in Q:
                          q.put(friend[o][0])
                          Q.append(friend[o][0])
          now = q.get()
          if (now not in F) and (now not in B):
              count += 1
      suggestions.append(count)
  
  print(" ".join(map(str, suggestions)))
  
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"}]