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: Wrong Answer)

N, M, K = [int(i) for i in input().split()]
F = [[int(i) for i in input().split()] for _ in range(M)]
B = [[int(i) for i in input().split()] for _ in range(K)]
clusters = []
cluster_label = [-1 for _ in range(N+1)]
friends = [[] for _ in range(N+1)]
blocks = [[] for _ in range(N+1)]
def make_cluster(i,j):
  cluster_label[i] = len(clusters)
  cluster_label[j] = len(clusters)
  clusters.append([i,j])
def search_cluster(i):
  return cluster_label[i]
def add_cluster(i, n):
  cluster_label[i] = n
  clusters[n].append(i)
def join_clusters(n, m):
  tmp = min(n, m)
  m = max(n, m)
  n = tmp
  for mem in clusters[m]:
    cluster_label[mem] = n
  clusters[n].extend(clusters[m])
  clusters[m] = -1
for i,j in F:
  p, q = search_cluster(i), search_cluster(j)
  friends[i].append(j)
  friends[j].append(i)
  if(p+q==-2):
    make_cluster(i, j)
  elif(p-q==0):
    continue
  elif(p==-1):
    add_cluster(i, q)
  elif(q==-1):
    add_cluster(j, p)
  else:
    join_clusters(p, q)
for i,j in B:
  blocks[i].append(j)
  blocks[j].append(i)
clusters.append([-1])
[print(len([j for j in clusters[cluster_label[i]] if((j not in friends[i])&(j not in blocks[i])&(j!=i))])) for i in range(1,N+1)]