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: Accepted)

from collections import deque
import sys
input = sys.stdin.readline
n, m, k = map(int, input().split())
friend = [[] for _ in range(n)]
for _ in range(m):
    a, b = map(int, input().split())
    friend[a-1].append(b-1)
    friend[b-1].append(a-1)
check = [0]*n
union = [-1]*n
union[0] = 0
for i in range(n):
    if check[i]: continue
    que = deque([i])
    while que:
        u = que.popleft()
        for v in friend[u]:
            if check[v]: continue
            union[v] = i
            check[v] = 1
            que.append(v)
            
brock = [[] for _ in range(n)]
for _ in range(k):
    c, d = map(int, input().split())
    if union[c-1] == union[d-1]:
        brock[c-1].append(d-1)
        brock[d-1].append(c-1)
        
ans = [0]*n
check = [0]*n
for i in range(n):
    if check[i] or union[i]==-1: continue
    que = deque([i])
    for j in range(i+1, n):
        if union[j] == union[i]: que.append(j)
    res = len(que)
    while que:
        u = que.popleft()
        check[u] = 1
        ans[u] = res-len(friend[u])-len(brock[u])-1
print(*ans)