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: Time Limit Exceeded)

from collections import deque

n,m,k = list(map(int, input().split()))


friendship = {}
for i in range(m):
    a, b = list(map(int, input().split()))
    if a in friendship.keys():
        friendship[a].append(b)
    else:
        friendship[a] = [b]
        
    if b in friendship.keys():
        friendship[b].append(a)
    else:
        friendship[b] = [a]
        
block = {}
for j in range(k):
    c, d = list(map(int, input().split()))
    if c in block.keys():
        block[c].append(d)
    else:
        block[c] = [d]
        
    if d in block.keys():
        block[d].append(c)
    else:
        block[d] = [c]
        
ans = [0] * n
for i in range(n):
    person = i+1
    checked_person = set()
    checked_person.add(person)
    if person in friendship.keys():
        candidate = deque(friendship[person])
        while len(candidate)>0:
            c = deque.popleft(candidate)
            checked_person.add(c)
            if c not in friendship[person]:
                if person in block.keys():
                    if c not in block[person]:
                        ans[i] += 1
                else:
                    ans[i] += 1
            if c in friendship.keys():
                for p in friendship[c]:
                    if p not in checked_person and p not in candidate:
                        candidate.append(p)
    else:
        continue
                
answer = ''
for i in range(len(ans)):
    answer += str(ans[i])
    if i != n-1:
        answer += ' '
            
print(answer)