Problem:
Score : 400 points

Problem 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?

Constraints

All values in input are integers.

2 ≤ N ≤ 10^5

0 \leq M \leq 10^5

0 \leq K \leq 10^5

1 \leq A_i, B_i \leq N

A_i \neq B_i

1 \leq C_i, D_i \leq N

C_i \neq D_i

(A_i, B_i) \neq (A_j, B_j) (i \neq j)

(A_i, B_i) \neq (B_j, A_j)

(C_i, D_i) \neq (C_j, D_j) (i \neq j)

(C_i, D_i) \neq (D_j, C_j)

(A_i, B_i) \neq (C_j, D_j)

(A_i, B_i) \neq (D_j, C_j)

Input

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

Output

Print the answers in order, with space in between.

Sample Input 1

4 4 1
2 1
1 3
3 2
3 4
4 1

Sample Output 1

0 1 0 1

There is a friendship between User 2 and 3, and between 3 and 4. Also, there is no friendship or blockship between User 2 and 4. Thus, User 4 is a friend candidate for User 2.

However, neither User 1 or 3 is a friend candidate for User 2, so User 2 has one friend candidate.

Sample Input 2

5 10 0
1 2
1 3
1 4
1 5
3 2
2 4
2 5
4 3
5 3
4 5

Sample Output 2

0 0 0 0 0

Everyone is a friend of everyone else and has no friend candidate.

Sample Input 3

10 9 3
10 1
6 7
8 2
2 5
8 4
7 3
10 9
6 4
5 8
2 6
7 5
3 1

Sample Output 3

1 3 5 4 3 3 3 3 1 0

Solution in Python (3.4.3) (judge verdict: Accepted)

from sys import setrecursionlimit

setrecursionlimit(3 * 10 ** 5)

def main():
  n, m, k = map(int, input().split(' '))
  nodes = tuple([([], set()) for i in range(n)])
  for _ in range(m):
    a, b = sorted(map(lambda x: int(x) - 1, input().split()))
    nodes[a][0].append(b)
    nodes[b][0].append(a)

  for _ in range(k):
    c, d = sorted(map(lambda x: int(x) - 1, input().split()))
    nodes[c][1].add(d)
    nodes[d][1].add(c)

  total_done_set = set()

  def dfs(node_id, done_set):
    if node_id in done_set: return 0

    total_done_set.add(node_id)
    done_set.add(node_id)

    size = 1
    for child_id in nodes[node_id][0]:
      size += dfs(child_id, done_set)

    return size


  ans_list = [-1] * n

  def dfs2(node_id, connected_ids, total_len):
    children, block_id_set = nodes[node_id]
    ans_list[node_id] = total_len - len(list(filter(
      lambda block_id: block_id in connected_ids, block_id_set
    ))) - 1

    except_num = 0
    for child_id in children:
      if child_id not in block_id_set:
        except_num += 1

      if ans_list[child_id] == -1:
        dfs2(child_id, connected_ids, total_len)

    ans_list[node_id] -= except_num


  for i in range(n):
    if i not in total_done_set:
      done_set = set()
      dfs(i, done_set)
      dfs2(i, done_set, len(done_set))

  print(*ans_list, sep=" ")


main()