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, :size
  
  def initialize(n)
    @parent = []
    @size = []
    n.times do |i|
      @parent[i] = i
      @size[i] = 1
    end
  end

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

  def merge(x,y)
    x,y = root(x),root(y)
    return if x==y

    x,y = y,x if @size[x] < @size[y]
    @size[x] += @size[y]
    @parent[y] = x
  end
  
  def same(x,y)
    return root(x) == root(y)
  end
  
  def size(x)
    return @size[root(x)]
  end
end

n,m,k = gets.split.map(&:to_i)
uf = UnionFind.new(n+1)
friendCnt = [0]*(n+1)
blockCnt = [0]*(n+1)
m.times do
  a,b = gets.split.map(&:to_i) 
  uf.merge(a,b)
  friendCnt[a] += 1
  friendCnt[b] += 1
end
k.times do
  c,d = gets.split.map(&:to_i)
  if uf.same(c,d)
    blockCnt[c] += 1
    blockCnt[d] += 1
  end
end

1.upto(n) do |i|
  print uf.size(i) - friendCnt[i] - blockCnt[i] - 1
  print i == n ? "\n" : " "
end