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)

used = [];
w = [];

def dfs(v, num):
    used[v] = num;
    for x in w[v]:
        if used[x] == 0:
            dfs(x, num);


n, m, k = (int(i) for i in input().split());

for i in range(n):
    w.append([]);

for j in range(m):
    u, v = (int(i) for i in input().split());
    u -= 1;
    v -= 1;
    w[u].append(v);
    w[v].append(u);

block = [];
for j in range(k):
    u, v = (int(i) for i in input().split());
    u -= 1;
    v -= 1;
    block.append([u, v]);

for i in range(n):
    used.append(0);

cnt_comp = 0;
for i in range(n):
    if used[i] == 0:
        cnt_comp += 1;
        dfs(i, cnt_comp);

cnt = [0 for i in range(cnt_comp)];

for i in range(n):
    cnt[used[i] - 1] += 1;

res = [0 for i in range(n)];
for i in range(n):
    res[i] = cnt[used[i] - 1] - 1;
#print(res);
for x in block:
    if used[x[0]] == used[x[1]]:
        res[x[0]] -= 1;
        res[x[1]] -= 1;
#print(res);
for i in range(n):
    for v in w[i]:
        if used[v] == used[i]:
            res[i] -= 1;

#print(cnt);

for i in res:
    print(i, end=" ");