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 (2.7.1) (judge verdict: Wrong Answer)

class UnionFind
    attr_reader :parent

    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+1)
ret = Array.new(N+1, 0)
M.times do
    a,b = gets.split.map(&:to_i)
    uf.unite(a,b)
    ret[a] -= 1
    ret[b] -= 1
end
K.times do
    c,d = gets.split.map(&:to_i)
    if uf.same_parent?(c,d)
        ret[c] -= 1
        ret[d] -= 1
    end
end
counts = Array.new(N+1,0)
uf.parent.group_by {|e| e}.each do |k,v|
    counts[k] = v.length
end
1.upto(N) do |i|
    ret[i] += counts[uf.get_parent(i)] - 1
end

puts ret[1..-1].join(' ')