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)

# 2020-05-22
from collections import deque


def main():
    n, m, k = [int(x) for x in input().split()]
    friends = [set() for _ in range(n)]
    blocks = [set() for _ in range(n)]
    for _ in range(m):
        a, b = [int(x) - 1 for x in input().split()]
        friends[a].add(b)
        friends[b].add(a)
    for _ in range(k):
        a, b = [int(x) - 1 for x in input().split()]
        blocks[a].add(b)
        blocks[b].add(a)

    ansers = []
    ansers_append = ansers.append
    nets = set()
    visited = set()
    visited_add = visited.add
    for person in range(n):
        if person in visited:
            # Already marked in one of the nets.
            for candidate_net in nets:
                if person in candidate_net:
                    net = candidate_net
                    break
        else:
            stack = deque()
            stack_append = stack.append
            stack_pop = stack.pop
            stack_append(person)
            net = {person}
            net_add = net.add
            while stack:
                new = stack_pop()
                for adj in friends[new]:
                    if adj in net:
                        continue
                    stack_append(adj)
                    net_add(adj)
            nets.add(tuple(net))

        # 友達をたどるとブロック対象に着くことがある。積集合 &。
        ans = len(net) - len(friends[person]) - len(net & blocks[person]) - 1
        ansers_append(ans)
        visited_add(person)
    print(*ansers)
    return


if __name__ == '__main__':
    main()