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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

def merge(a,b)
  root_a = root(a)
  root_b = root(b)

  if root_a == root_b then
    return
  end
  
  if root_a < root_b then
    root_a, root_b = root_b, root_a
  end
  $par[root_a] = root_b
end

def root(x)
  if $par[x] == x then
    return x
  else
    return $par[x] = root($par[x])
  end
end

N_M_K = gets.chomp
N = N_M_K.split(" ")[0].to_i
M = N_M_K.split(" ")[1].to_i
K = N_M_K.split(" ")[2].to_i

exep = Array.new(N){Array.new()}

$par = (0..N-1).reduce([]){|par, i|
  par.push(i)
}

for i in 0..M-1 
  a_b = gets.chomp
  a = a_b.split(" ")[0].to_i - 1
  b = a_b.split(" ")[1].to_i - 1
  exep[a].push(b)
  exep[b].push(a)
  merge(a,b)
end
  
for i in 0..K-1 
  a_b = gets.chomp
  a = a_b.split(" ")[0].to_i - 1
  b = a_b.split(" ")[1].to_i - 1
  if root(a) != root(b) then
    next
  end
  exep[a].push(b)
  exep[b].push(a)
end

(1..N-1).each do |i|
  root(i)
end

puts (0..N-1).reduce(""){|sum,i|
  sum += "#{$par.count($par[i]) - 1 - exep[i].count} "
}