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: Accepted)

n, m, k = map(int, input().split())
par = [-1 for _ in range(n)]

def find(x):
    if par[x] < 0:
        return x
    else:
        par[x] = find(par[x])  # 経路圧縮
        return par[x]

def same(x, y):
    return find(x) == find(y)

def unite(x, y):
    x = find(x)
    y = find(y)
    if x == y:
        return False
    # サイズの大きい木にサイズの小さい木をくっつける。親の直下に親をくっつける。
    # 親のサイズは負の値になっているので、符号に注意
    if par[x] < par[y]:
        par[x] += par[y]
        par[y] = x
    else:
        par[y] += par[x]
        par[x] = y
    return True

friends_num = [0] * 100005
for _ in range(m):
    num_from, num_to = map(int, input().split())
    num_from -= 1
    num_to -= 1
    friends_num[num_from] += 1
    friends_num[num_to] += 1
    unite(num_from, num_to)

blocks = [[] * n for i in range(n)]
for _ in range(k):
    num_from, num_to = map(int, input().split())
    num_from -= 1
    num_to -= 1
    blocks[num_from].append(num_to)
    blocks[num_to].append(num_from)

ans_list = []
for i in range(n):
    ans = -par[find(i)] - friends_num[i] - 1
    for b in blocks[i]:
        if same(i, b):
            ans -= 1
    ans_list.append(ans)

print(' '.join(map(str, ans_list)))