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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

import sys
sys.setrecursionlimit(10*9)

n,m,k = map(int,input().split())
parent_of = [-1 for _ in range(n)]
#print(parent_of)
def root_of(x):
	if parent_of[x] < 0:
		return x
	else:
		parent_of[x] = root_of(parent_of[x])
		return parent_of[x]

def unite(x,y):
	top_root_x = root_of(x)
	top_root_y = root_of(y)
	if top_root_x == top_root_y:
		return False
	if parent_of[top_root_x] > parent_of[top_root_y]:
		top_root_x,top_root_y = top_root_y,top_root_x
	#print(parent_of[top_root_y])
	parent_of[top_root_x] += parent_of[top_root_y]
	parent_of[top_root_y] = top_root_x
	return True

def size_of(x):
		return -parent_of[root_of(x)]

def is_same(x,y):
	return root_of(x) == root_of(y)

block_list = [0] * n
friend_list = [0] * n
for _ in range(m):
	a,b = map(int,input().split())
	a -= 1
	b -= 1
	friend_list[a] +=1
	friend_list[b] +=1
	unite(a,b)

for _ in range(k):
	c,d = map(int,input().split())
	c -= 1
	d -= 1
	if root_of(c) == root_of(d):
		block_list[c] += 1
		block_list[d] += 1
answer_array = [-1] * n
for i in range(n):
	answer_array[i] = size_of(i) - friend_list[i] - block_list[i] - 1
	#print(size_of(i))
print(*answer_array)