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)

n, m, k = gets.split.map(&:to_i)

relations = [*(0...n)].map do |num|
  tmp = Array.new(n, nil)
  tmp[num] = 'N'
  tmp
end

m.times do
  a, b = gets.split.map(&:to_i)
  relations[a-1][b-1] = 'F'
  relations[b-1][a-1] = 'F'
end

k.times do
  c, d = gets.split.map(&:to_i)
  relations[c-1][d-1] = 'B'
  relations[d-1][c-1] = 'B'
end

def relation?(relations, a, b, queue = [])
  return relations[a][b] unless relations[a][b].nil?

  a_frends = relations[a].map.with_index { |e,i| ['F', 'C'].include?(e) ? i : nil }.compact
  a_frends.each { |friend| queue << [[a], friend, b] }

  while queue.size.positive?
    arr, friend, b = queue.pop

    if ['F', 'C'].include?(relations[friend][b])
      arr.each do |aa|
        relations[aa][b] = 'C'
        relations[b][aa] = 'C'
      end
      break
    end

    friend_friends = relations[friend].map.with_index { |e,i| ['F', 'C'].include?(e) ? i : nil }.compact
    friend_friends.each { |friend_friend| queue << [arr << friend, friend_friend, b] }
  end

  relations[a][b] = 'N' if relations[a][b].nil?

  relations[a][b]
end
# puts relation?(relations, 1,3)

ans = relations.map.with_index do |one, a|
  count = 0
  one.each_with_index do |rel, b|
    count += 1 if relation?(relations, a, b) == 'C'
  end
  # p relations
  count
end

# p relations

puts ans.join(' ')