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 (7.3.0) (judge verdict: Wrong Answer)

#!/usr/bin/env python3
import sys
import collections as cl


def II(): return int(sys.stdin.readline())


def MI(): return map(int, sys.stdin.readline().split())


def LI(): return list(map(int, sys.stdin.readline().split()))


def main():
    N, M, K = MI()
    friends = [[] for i in range(N)]
    blocks = [[] for i in range(N)]

    for _ in range(M):
        a, b = MI()
        a -= 1
        b -= 1
        friends[a].append(b)
        friends[b].append(a)

    for _ in range(K):
        a, b = MI()
        a -= 1
        b -= 1
        blocks[a].append(b)
        blocks[b].append(a)

    groups = []
    flag = [-1] * N
    group = 0
    for i in range(N):
        if flag[i] != -1:
            continue
        deque = cl.deque(friends[i])
        member_tmp = []
        while len(deque) > 0:
            target = deque.popleft()
            friends_i = friends[target]
            for frend in friends_i:
                if flag[frend] != -1:
                    continue
                member_tmp.append(frend)
                flag[frend] = group
                deque.append(frend)
        groups.append(member_tmp)
        group += 1

    ans = []

    for i in range(len(groups)):
        groups[i] = set(groups[i])

    for i in range(N):
        i_group = flag[i]
        num_group = len(groups[i_group] - set(friends[i]) - set(blocks[i])) - 1
        ans.append(num_group)

    print(" ".join(map(str, ans)))


main()