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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

N, M, K = map(int, input().split())
friend = [list(map(int, input().split())) for i in range(M)]
block = [list(map(int, input().split())) for i in range(K)]

result = []
FG = [[] for i in range(N + 1)]
BG = [[] for i in range(N + 1)]
 
for i in range(M):
    a = friend[i][0]
    b = friend[i][1]
    FG[a].append(b)
    FG[b].append(a)
 
for i in range(K):
    a = block[i][0]
    b = block[i][1]
    BG[a].append(b)
    BG[b].append(a)
    
result = [0] * (N + 1)
flag = [0] * (N + 1)
point = 1


while point <= N:
    group = []
    if flag[point] == 0:# 集合を探索していない場合
        group.append(point)
        flag[point] = 1
        p = 0
        while p < len(group):
            now = group[p]            
            if len(FG[now]) != 0:# 友達が1人以上いる場合、友達を集合に加える
                for i in range(len(FG[now])):
                    if flag[FG[now][i]] == 0:
                        group.append(FG[now][i])
                        flag[FG[now][i]] = 1
            p += 1
        S = len(group)
        for j in range(S):
            m = group[j]
            FB = set(FG[m] + BG[m])
            SG = set(group)
            fc = list(SG - FB)
            result[m] = len(fc) - 1
    point += 1
            
print(*result[1:])