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

  def root(x)
    @parent[x] < 0 ? x : @parent[x] = root(@parent[x])
  end

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

  def unite(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
    return true
  end

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

require 'set'; require 'prime'
INF=Float::INFINITY; MOD=10**9+7
n,m,k = gets.chomp.split.map(&:to_i)

uf = UnionFind.new(n+1)
fh = {}
m.times do
  a, b = gets.chomp.split.map(&:to_i)
  a-=1
  b-=1
  uf.unite(a,b)

  fh[a] ||= Set.new
  fh[a].add b
  fh[b] ||= Set.new
  fh[b].add a
end

h = {}
k.times do
  c,d = gets.chomp.split.map(&:to_i)
  c-=1
  d-=1
  h[c] ||= Set.new
  h[c].add d
  h[d] ||= Set.new
  h[d].add c
end

parents = Array.new(n)
frient_sets = []
n.times do |i|
  parent = uf.root(i)
  parents[i] = parent
  frient_sets[parent] ||= Set.new
  frient_sets[parent].add i
end

ans = Array.new(n)
n.times do |i|
  parent = parents[i]
  frient_set = frient_sets[parent]
  # p [i, frient_set, fh[i], h[i]]
  frient_set -= fh[i] unless fh[i].nil?
  frient_set -= h[i] unless h[i].nil?
  ans[i] = frient_set.size - 1
end

puts ans.join(' ')