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)

graph = Array.new(n) { -1 }

friends = []
m.times do
  friends << gets.split.map(&:to_i)
end

enemies = []
k.times do
  enemies << gets.split.map(&:to_i)
end

def get_friends(friends, i)
  (friends.select { |f| f[0] == i}.map { |f| f[1] } + friends.select { |f| f[1] == i}.map { |f| f[0] }).uniq
end

def get_enemies(enemies, i)
  (enemies.select { |f| f[0] == i}.map { |f| f[1] } + enemies.select { |f| f[1] == i}.map { |f| f[0] }).uniq
end

def dfs(graph, friends, i, label)
  return unless graph[i-1] == -1
  # label += 1 if graph.select { |s| s == label }.count == 0
  
  graph[i-1] = label
  target = get_friends(friends, i)
  target.each do |f|
    dfs(graph, friends, f, label)
  end
end

def get_candidates(graph, friends, enemies, i)
  all = graph.select { |s| s == graph[i-1] }.count
  all = all - 1 - get_friends(friends, i).count
  all - get_enemies(enemies, i).select { |s| graph[i-1] == graph[s-1]}.count
end

n.times do |i|
  dfs(graph, friends, i+1, i)
end

n.times do |i|
  print get_candidates(graph, friends, enemies, i+1), " "
end
print "\n"