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: Time Limit Exceeded)

# -*- coding: utf-8 -*-
"""
Created on Mon Mar  2 06:12:32 2020

@author: naoki
"""
from collections import deque
# =============================================================================
# with open('C:/Users/naoki/Desktop/Atcoder/input.txt') as f:
#     N ,M ,K = map(int,f.readline().split())
#     A = [list(map(int,f.readline().split())) for _ in range(M)]
#     C = [list(map(int,f.readline().split())) for _ in range(K)]
# =============================================================================

N ,M ,K = map(int,input().split())
A = [list(map(int,input().split())) for _ in range(M)]
C = [list(map(int,input().split())) for _ in range(K)]

seen = {i:False for i in range(1,N+1)}
label = {i:0 for i in range(1,N+1)}
todo = deque()
All = 0
num = 0
while False in seen.values():
    num+=1
    H = [k for k,v in seen.items() if v == False][0]
    todo.append(H)
    seen[H] = True
    group = set()
    while len(todo) > 0:
        current = todo.pop()
        label[current] = num
        group.add(current)
        for connection in A:
            if connection[0] in group and connection[1] in group:
                continue
            elif connection[0] in group and seen[connection[1]] == False:
                todo.append(connection[1])
                seen[connection[1]] = True
            elif connection[1] in group and seen[connection[0]] == False:
                todo.append(connection[0])
                seen[connection[0]] = True

minus = {i:0 for i in range(1,N+1)}
for pair in A:
    minus[pair[0]] -= 1
    minus[pair[1]] -= 1
for pair in C:
    if label[pair[0]] == label[pair[1]]:
        minus[pair[0]] -= 1
        minus[pair[1]] -= 1
accompany = {i:0 for i in range(1,num+1)}
for i in range(1,N+1):
    accompany[label[i]] += 1

for i in range(1,N+1):
    print(accompany[label[i]] + minus[i]-1)
    

# =============================================================================
# A = {5:[5,4],4:[2,3]}
# A[5].append(6)
# del A[5][0]
# =============================================================================

# =============================================================================
#     block = 0
#     for block in C:
#         if block[0] in group and block in group:
#             cone_num -= 2
# =============================================================================