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: Accepted)

class UnionFind
  attr_accessor :parent

  def initialize(size)
    @parent = Array.new(size, -1)
  end

  def root(x)
    if @parent[x] < 0
      x
    else
      @parent[x] = root(@parent[x])
    end
  end

  def same_parent?(x, y)
    root(x) == root(y)
  end

  def merge(x, y)
    x_root = root(x)
    y_root = root(y)

    return false if x_root == y_root

    if @parent[x_root] > @parent[y_root]
      x_root, y_root = y_root, x_root
    end

    @parent[x_root] += @parent[y_root]
    @parent[y_root] = x_root

    true
  end

  def size(x)
    -@parent[root(x)]
  end
end

n, m, k = gets.chomp.split(' ').map(&:to_i)
arr = Array.new(n)
union = UnionFind.new(n)
n.times do |i|
  arr[i] = {}
end
m.times do
  a, b = gets.chomp.split(' ').map { |i| i.to_i - 1 }
  union.merge(a, b)
  arr[a][b] = true
  arr[b][a] = true
end
k.times do
  a, b = gets.chomp.split(' ').map { |i| i.to_i - 1 }
  if union.same_parent?(a, b)
    arr[a][b] = true
    arr[b][a] = true
  end
end
res = []
n.times do |i|
  res << union.size(i) - arr[i].size - 1
end
puts res.join(' ')