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 = map(int, input().split())
AB = [list(map(int, input().split())) for i in range(M)]
CD = [list(map(int, input().split())) for i in range(K)]

frends = [set() for i in range(N)]
not_frends_koho = [set() for i in range(N)]
frends_koho = []
people = set(range(N))

for relF in AB:
    frends[relF[0]-1].add(relF[1]-1)
    frends[relF[1]-1].add(relF[0]-1)

for relF in AB:
    not_frends_koho[relF[0]-1].add(relF[1]-1)
    not_frends_koho[relF[1]-1].add(relF[0]-1)

for relB in CD:
    not_frends_koho[relB[0]-1].add(relB[1]-1)
    not_frends_koho[relB[1]-1].add(relB[0]-1)
  
for i in range(len(not_frends_koho)):
    frends_koho.append((people - not_frends_koho[i]) - set([i]))
    
memo = []

def search(idx, goal, memo_str):
    if str(goal) in memo_str and memo_str not in memo and memo_str[-1] == str(goal):
        memo.append(memo_str)
        return
    
    if str(idx) in memo_str:
        return
    
    memo_str += str(idx)
    
    for num in frends[idx]:
        search(num, goal, memo_str)


for start in range(len(frends_koho)):
    if frends_koho[start]:
        for end in frends_koho[start]:
            search(start, end, '')

memo = set([num[0] + num[-1] for num in memo])

ans = [0 for i in range(N)]
for i in memo:
    ans[int(i[0])] += 1

for i in ans:print(i, end=' ')