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, Counter
  import sys
  
  N_MAX = 100000 + 5
  sys.setrecursionlimit(N_MAX)
  
  N, M, K = map(int, sys.stdin.readline().rstrip().split())
  
  fr = [[] for _ in range(N)]  # 友達関係
  for _ in range(M):
      a, b = map(int, sys.stdin.readline().rstrip().split())
      fr[a - 1].append(b - 1)
      fr[b - 1].append(a - 1)
  
  bl = [[] for _ in range(N)]  # ブロック関係
  for _ in range(K):
      a, b = map(int, sys.stdin.readline().rstrip().split())
      bl[a - 1].append(b - 1)
      bl[b - 1].append(a - 1)
  
  
  gr = [0] * N  # 友達候補グループ
  
  
  # ## BFS ## #
  def bfs(u, num):
  
      q = deque()
      q.append(u)
      while q:
          u = q.popleft()
          for v in fr[u]:
              if gr[v] == 0:  # state を確認
                  gr[v] = num  # state を変更
                  q.append(v)
  
  
  num = 1
  for i in range(N):
      if gr[i] == 0:
          bfs(i, num)
          num += 1
  
  # print(gr)
  
  gr_num = Counter(gr)  # グループの人数を数えておく
  
  
  # グループの中から、友達でも
  ans = []
  for i in range(N):
      kouho_suu = gr_num[gr[i]]
      for f in fr[i]:
          if gr[i] == gr[f]:
              kouho_suu -= 1
      for b in bl[i]:
          if gr[i] == gr[b]:
              kouho_suu -= 1
      kouho_suu -= 1
  
      ans.append(kouho_suu)
  
  print(*ans)
  
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"}]