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
  def initialize(n); @data = Array.new(n,-1); end
  def find(a); @data[a] < 0 ? a : @data[a] = find(@data[a]); end
  def same?(a,b); find(a) == find(b); end
  def size(a); -@data[find(a)]; end
  def unite(a,b)
    a = find(a); b = find(b)
    return if a == b
    a,b = b,a if @data[a] > @data[b]
    @data[a] += @data[b]; @data[b] = a
  end
end

# UnionFind木で連結成分の要素数を求める
# 友人、ブロックの隣接リスト作成
# 人別(N),友人数(M)で二重ループしながらUF参照
# Mが疎なので計算量はO(M)、ブロックについてもO(K)

n,m,k = gets.split.map &:to_i
UF = UnionFind.new(n)
FR = Array.new(n){[]}
BL = Array.new(n){[]}

m.times do
  a,b = gets.split.map &:to_i
  a -= 1; b -= 1
  UF.unite(a,b)
  FR[a] << b
  FR[b] << a
end

k.times do
  a,b = gets.split.map &:to_i
  a -= 1; b -= 1
  BL[a] << b
  BL[b] << a
end

ans = n.times.map do |i|
  cnt = UF.size(i)
  FR[i].each do |f|
    cnt -= 1 if UF.same?(i,f)
  end
  BL[i].each do |b|
    cnt -= 1 if UF.same?(i,b)
  end
  cnt -= 1
end

puts ans.join(" ")