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)

# function
def another(rel, per)
  return rel.reject{|i|i==per}[0]
end

def search(current, from)
  if $complete.include?(current) then
    return
  end
    
  $complete.push(current)
  
  newSelec = $fri.select{|i|
    i.include?(current)
  }
    
  if newSelec.count <= 0 then
    return
  end
    
  newSelec.each do |i|
    nextRel = another(i, current)

    fromTo  = [from,nextRel]
    fromTo = [fromTo.min, fromTo.max]
	
    if nextRel == from  then
      next
    elsif current == from then
    elsif $blk.include?(fromTo) then
    elsif $fri.include?(fromTo) then
    elsif $counted.include?(fromTo) == false then
      $cnt[from - 1]    = $cnt[from - 1] + 1
      $cnt[nextRel - 1] = $cnt[nextRel - 1] + 1
      $counted.push(fromTo)
    end
    search(nextRel, from)
  end
end

def logarr(arr)
 return "[#{arr[0]},#{arr[1]}]"
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

$cnt = Array.new(N).fill(0)
$fri = Array.new(M)
$blk = Array.new(K)

for i in 0..M-1 
  a_b = gets.chomp
  hoge = [a_b.split(" ")[0].to_i, a_b.split(" ")[1].to_i]
  $fri[i] = [hoge.min, hoge.max]
end
  
for i in 0..K-1 
  a_b = gets.chomp
  hoge = [a_b.split(" ")[0].to_i, a_b.split(" ")[1].to_i]
  $blk[i] = [hoge.min, hoge.max]
end

$counted = []
for i in 1..N
  $complete  = []
  search(i, i)
end
  
puts $cnt
 
exit