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)

require "set"
n,m,k = gets.chomp.split(" ").map(&:to_i)

ab_arr = m.times.map{gets.chomp.split(" ").map(&:to_i)}
cd_arr = k.times.map{gets.chomp.split(" ").map(&:to_i)}

f_hash = {}
ab_arr.each do |(a, b)|
  f_hash[a] ||= Set.new
  f_hash[a] << b
  f_hash[b] ||= Set.new
  f_hash[b] << a
end

b_hash = {}
cd_arr.each do |(c, d)|
  b_hash[c] ||= Set.new
  b_hash[c] << d
  b_hash[d] ||= Set.new
  b_hash[d] << c
end

result = []
(1..n).each do |i|

  arr = (f_hash[i]&.to_a || [])
  flags = Set.new

  fs = (f_hash[i] || Set.new)
  bs = (b_hash[i] || Set.new)
  f_set = Set.new
  while arr.length > 0 do
    a = arr.pop
    flags << a

    ff = (f_hash[a] || Set.new).to_a

    ff.each do |f|
      if i != f && !fs.include?(f) && !bs.include?(f) then
        f_set << f
      end
      if !flags.include?(f) then
        arr << f
      end
    end
  end
  result << f_set.length
end

puts result.join(" ")