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: used = []
  w = []
  
  
  def dfs(v, num):
      used[v] = num
      for x in w[v]:
          if used[x] == 0:
              dfs(x, num)
  
  
  n, m, k = (int(i) for i in input().split())
  
  for i in range(n):
      w.append([])
  
  for j in range(m):
      u, v = (int(i) for i in input().split())
      u -= 1
      v -= 1
      w[u].append(v)
      w[v].append(u)
  
  block = []
  for j in range(k):
      u, v = (int(i) for i in input().split())
      u -= 1
      v -= 1
      block.append([u, v])
  
  for i in range(n):
      used.append(0)
  
  # cnt_comp = 0;
  # for i in range(n):
  #     if used[i] == 0:
  #         cnt_comp += 1;
  #         dfs(i, cnt_comp);
  
  # cnt = [0 for i in range(cnt_comp)];
  #
  # for i in range(n):
  #     cnt[used[i] - 1] += 1;
  #
  # res = [0 for i in range(n)];
  # for i in range(n):
  #     res[i] = cnt[used[i] - 1] - 1;
  # #print(res);
  # for x in block:
  #     if used[x[0]] == used[x[1]]:
  #         res[x[0]] -= 1;
  #         res[x[1]] -= 1;
  # #print(res);
  # for i in range(n):
  #     for v in w[i]:
  #         if used[v] == used[i]:
  #             res[i] -= 1;
  #
  # #print(cnt);
  #
  # for i in res:
  #     print(i, end=" ");
  
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"}]