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: Time Limit Exceeded)

import sys
from collections import deque
input = sys.stdin.readline

def main():
    
    N, M, K = map(int, input().split())
    G = {v: deque([]) for v in range(N)}
    G_original = {v: deque([]) for v in range(N)}
    for _ in range(M):
        a, b = map(int, input().split())
        a -= 1
        b -= 1
        G[a].append(b)
        G[b].append(a)
        G_original[a].append(b)
        G_original[b].append(a)

    Block = {v: deque([]) for v in range(N)}
    for _ in range(K):
        c, d = map(int, input().split())
        c -= 1
        d -= 1
        Block[c].append(d)
        Block[d].append(c)

    # seen = [0]*(N)
    stack = []
    color = {v: -1 for v in range(N)}
    col2v = {color: set() for color in range(N+4)}
    
    def dfs(j, col):
        if color[j] != -1:
            return
        # seen[j] = 1
        stack.append(j)
        color[j] = col
        col2v[col].add(j)
        while stack:
            s = stack[-1]
            if not G[s]:
                stack.pop()
                continue
            g_NO  = G[s].popleft()
            if color[g_NO] != -1:
                continue
            color[g_NO] = col
            col2v[col].add(g_NO)
            # color[g_NO] = 1
            stack.append(g_NO)
        
    c = 0
    for start in range(N):
        if color[start]!=-1:
            continue
        dfs(start, c)
        c += 1
    
    ans = []
    for v in range(N):
        kouho = col2v[color[v]]
        # print(kouho)
        block = Block[v]
        # print(block)
        friend = G_original[v]
        # print(friend)
        tmp = kouho.difference(block).difference(friend)
        ans.append(len(tmp)-1)
        
        # print("##############")
    print(*ans)
main()