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: def s0():
      return input()
  
  
  def s1():
      return input().split()
  
  
  def s2(n):
      return [input() for x in range(n)]
  
  
  def s3(n):
      return [input().split() for _ in range(n)]
  
  
  def s4(n):
      return [[x for x in s] for s in s2(n)]
  
  
  def n0():
      return int(input())
  
  
  def n1():
      return [int(x) for x in input().split()]
  
  
  def n2(n):
      return [int(input()) for _ in range(n)]
  
  
  def n3(n):
      return [[int(x) for x in input().split()] for _ in range(n)]
  
  
  def t3(n):
      return [tuple(int(x) for x in input().split()) for _ in range(n)]
  
  
  def p0(b, yes="Yes", no="No"):
      print(yes if b else no)
  
  
  # from sys import setrecursionlimit
  # setrecursionlimit(1000000)
  # from collections import Counter,deque,defaultdict
  # import itertools
  # import math
  import networkx as nx
  
  # from bisect import bisect_left,bisect_right
  # from heapq import heapify,heappush,heappop
  n, m, k = n1()
  AB = n3(m)
  CD = n3(k)
  
  G = nx.Graph()
  
  G.add_edges_from(AB)
  
  friend = {i: [] for i in range(1, n + 1)}
  for i, j in AB:
      friend[i].append(j)
      friend[j].append(i)
  
  block = {i: [] for i in range(1, n + 1)}
  for i, j in CD:
      block[i].append(j)
      block[j].append(i)
  
  ans = []
  for i in range(1, n + 1):
      g = nx.single_source_shortest_path(G, source=i)
      f_kouho = set(g.keys())
      tmp = -1
      for j in f_kouho:
          if (not j in friend[i]) and (not j in block[i]):
              tmp += 1
      ans.append(tmp)
  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"}]