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: Runtime Error)

def dfs(cur_node, candidates, visited, level, cnt):

  while len(candidates) > 0:
    node, level = candidates.pop(0)

    if visited[node]: continue
    visited[node] = True

    if level > 0: cnt += 1

    for n in adj_list[node][FRIEND]:
      if n in adj_list[cur_node][BLOCK]:
        candidates.append((n, 0))
      if visited[n] == False:
        candidates.append((n, level+1))

  return cnt

# -----------------------------------------

FRIEND, BLOCK = 0, 1

N, F, B = list(map(int, input().split()))
adj_list = defaultdict(dict)
for i in range(1, N+1):
  adj_list[i][0] = set()
  adj_list[i][1] = set()

for _ in range(F):
  u, v = list(map(int, input().split()))
  adj_list[u][FRIEND].add(v)
  adj_list[v][FRIEND].add(u)

for _ in range(B):
  u, v = list(map(int, input().split()))
  adj_list[u][BLOCK].add(v)
  adj_list[v][BLOCK].add(u)

ans = []

for i in range(1, N+1):
  visited = [False] * (N+1)
  visited[i] = True
  candidates = []
  for node in adj_list[i][FRIEND]:
    # node, level
    candidates.append((node, 0))

  cnt = 0
  cnt = dfs(i, candidates, visited, 0, 0)
  ans.append(cnt)

print(" ".join([str(i) for i in ans]))