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)

require "set"
N, M, K = gets.chomp.split(" ").map(&:to_i)
F = {}
B = {}
M.times do |i|
  a, b = gets.chomp.split(" ").map(&:to_i)
  F[a] ||= Set.new
  F[a] << b
  F[b] ||= Set.new
  F[b] << a
end
K.times do |i|
  c, d = gets.chomp.split(" ").map(&:to_i)
  B[c] ||= Set.new
  B[c] << d
  B[d] ||= Set.new
  B[d] << c
end
NG = {}
EMPTY = Set.new
(1..N).each do |i|
  NG[i] = (F[i] || EMPTY) + (B[i] || EMPTY)
  NG[i] << i
end

def search_friends(h, rest)
  visited = Set.new
  stack = []
  stack << h
  while !stack.empty?
    n = stack.pop
    next if visited.include?(n) || !rest.include?(n)
    visited << n
    next if F[n].nil?
    F[n].each {|v| stack << v}
  end
  visited
end

rest = (1..N).to_set
ans = Array.new(N, nil)
while !rest.empty?
  f = search_friends(rest.first, rest)
  f.each do |v|
    ans[v-1] = (f - NG[v]).size
  end
  rest -= f
end

puts ans.join(" ")