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: Runtime Error)

lines = []
while line = gets
    lines << line.chomp.split(' ').map(&:to_i)
end

N = lines[0][0]
M = lines[0][1]
K = lines[0][2]

friend = []
for i in 1..N do
  friend[i-1] = []
end
for i in 1..M do
  friend[lines[i][0]-1].push(lines[i][1])
  friend[lines[i][1]-1].push(lines[i][0])
end

block = []
for i in 1..N do
  block[i-1] = []
end
for i in 1..K do
  block[lines[M+i][0]-1].push(lines[M+i][1])
  block[lines[M+i][1]-1].push(lines[M+i][0])
end

finished = []
for i in 1..N do
  finished[i-1] = false
end

target = [1]
cluster = []
cluster[0] = []
num_cluster = 0
cluster[num_cluster].push(target[0])
while target.length != 0 do
  finished[target[0]-1] = true
  friend[target[0]-1].each do |i|
    if finished[i-1] == false then
      cluster[num_cluster].push(i)
      finished[i-1] = true
      target.insert(1,i)
    end
  end
  target.delete_at(0)
  if target.length == 0 then
    a = nil
    for i in 1..N do
      if finished[i-1] == false then
        a = i
        break
      end
    end
    if a != nil then
      target.push(a+1)
      num_cluster += 1
      cluster[num_cluster] = []
      cluster[num_cluster].push(target[0])
    end
  end
end

connected = []
for i in 1..N do
  connected[i-1] = []
end
for i in 1..cluster.length do
  cluster[i-1].each do |j|
    connected[j-1] = cluster[i-1]
  end
end
    
pot_friend = []
num_pot_friend = []
for i in 1..N do
  pot_friend[i-1] = []
  pot_friend[i-1] = connected[i-1] - friend[i-1] - block[i-1]- [i]
  num_pot_friend.push(pot_friend[i-1].length)
end
    
answer = ""
for i in 1..N do
  answer += num_pot_friend[i-1].to_s + " "
end
 
print answer