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: Runtime Error)

import sys
sys.setrecursionlimit(2147483647)
input = sys.stdin.readline
N, M, K = map(int, input().split())
graph = [[] for _ in range(N)]
block = [[] for _ in range(N)]
connected_components = []
count = 0
counts = [0 for _ in range(N)]
seen = [False for _ in range(N)]
def dfs(v, c, seen, origin):
    global count
    seen[v] = True
    for node in graph[v]:
        if seen[node] == True:
            continue
        if node not in block[origin]:
            count += 1
        dfs(node, c+1, seen, origin)
def dfs_connected(v, order):
    seen[v] = True
    connected_components[order].append(v)
    for node in graph[v]:
        if seen[node] == True:
            continue
        dfs_connected(node, order)
 
for Mi in range(M):
    a, b = map(int, input().split())
    graph[a-1].append(b-1)
    graph[b-1].append(a-1)
 
for Ki in range(K):
    a, b = map(int, input().split())
    block[a-1].append(b-1)
    block[b-1].append(a-1)
 
for v in range(N):
    if seen[v] == True:
        continue
    connected_components.append([])
    dfs_connected(v, v)
 
for components in connected_components:
    for component in components:
        seen = [False for _ in range(N)]
        count = 0
        dfs(component, count, seen, component)
        counts[component] = count-len(graph[component])
 
print(' '.join(map(str, counts)))