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 sys import stdin
  
  
  def getval():
      n, m, k = map(int, stdin.readline().split())
      a = [list(map(int, stdin.readline().split())) for i in range(m)]
      c = [list(map(int, stdin.readline().split())) for i in range(k)]
      return n, m, k, a, c
  
  
  def main(n, m, k, a, c):
      # Compute all members' friends
      friends = [[] for i in range(n)]
      for i in a:
          friends[i[0] - 1].append(i[1] - 1)
          friends[i[1] - 1].append(i[0] - 1)
  
      # Group all members in the isolated "friend groups"
      frgroups = []
      visited = [False for i in range(n)]
      for i in range(n):
          if visited[i]:
              continue
          q = [i]
          temp = [i]
          visited[i] = True
          while q:
              idx = q.pop(0)
              adj = friends[idx]
              for j in adj:
                  if visited[j]:
                      continue
                  visited[j] = True
                  q.append(j)
                  temp.append(j)
          frgroups.append(temp)
  
      # Number all members with their friend groups respectively
      groups = [-1 for i in range(n)]
      for i in range(len(frgroups)):
          for j in frgroups[i]:
              groups[j] = i
  
      # For each member, refer to the friend groups of blocked peope
      # Compute ans accordingly
      ans = [len(frgroups[groups[i]]) - 1 - len(friends[i]) for i in range(n)]
      for i in c:
          if groups[i[0] - 1] == groups[i[1] - 1]:
              ans[i[0] - 1] -= 1
              ans[i[1] - 1] -= 1
      s = str(ans[0])
      for i in range(1, n):
          s += " " + str(ans[i])
      print(s)
  
  
  if __name__ == "__main__":
      n, m, k, a, c = getval()
      main(n, m, k, a, c)
  
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"}]