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)

class UnionFind
  def initialize(size)
    @rank = Array.new(size, 0)
    @parent = Array.new(size, &:itself)
  end

  def unite(id_x, id_y)
    x_parent = get_parent(id_x)
    y_parent = get_parent(id_y)
    return if x_parent == y_parent

    if @rank[x_parent] > @rank[y_parent]
      @parent[y_parent] = x_parent
    else
      @parent[x_parent] = y_parent
      @rank[y_parent] += 1 if @rank[x_parent] == @rank[y_parent]
    end
  end

  def get_parent(id_x)
    @parent[id_x] == id_x ? id_x : (@parent[id_x] = get_parent(@parent[id_x]))
  end

  def same_parent?(id_x, id_y)
    get_parent(id_x) == get_parent(id_y)
  end
end

n,m,k=gets.split.map(&:to_i)
uf=UnionFind.new(n)
fr={}
m.times do
  a,b=gets.split.map(&:to_i)
  uf.unite(a-1, b-1)
  fr[a-1] ||= []
  fr[a-1] << b-1
  fr[b-1] ||= []
  fr[b-1] << a-1
end

bl={}
k.times do
  c,d=gets.split.map(&:to_i)
  bl[c-1] ||= []
  bl[c-1] << d-1
  bl[d-1] ||= []
  bl[d-1] << c-1
end
res = (0..(n-1)).to_a.map do |i|
  cnt=0
  n.times do |j|
    next if i == j
    next if bl[i]&.include?(j)
    next if fr[i]&.include?(j)
    cnt += 1 if uf.same_parent?(i, j)
  end
  cnt
end

puts res.join(' ')