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 LuaJIT (2.0.4) (judge verdict: Accepted)

local n, m, k = io.read("*n", "*n", "*n")

local edge = {}
local avoid = {}
for i = 1, n do
  edge[i] = {}
  avoid[i] = {}
end

for i = 1, m do
  local a, b = io.read("*n", "*n")
  edge[a][b], edge[b][a] = true, true
end
for i = 1, k do
  local c, d = io.read("*n", "*n")
  avoid[c][d], avoid[d][c] = true, true
end
local parent = {}
for i = 1, n do
  parent[i] = i
end

local function uf_findroot(idx)
  local idx_update = idx
  while parent[idx] ~= idx do
    idx = parent[idx]
  end
  while parent[idx_update] ~= idx do
    parent[idx_update], idx_update = idx, parent[idx_update]
  end
  return idx
end

for i = 1, n do
  local src = i
  local rs = uf_findroot(src)
  for dst, _u in pairs(edge[src]) do
    -- table.insert(tasks, dst)
    local rd = uf_findroot(dst)
    parent[rd], parent[dst] = rs, rs
  end
end
local groups = {}
for i = 1, n do
  local r = uf_findroot(i)
  if not groups[r] then
    groups[r] = 1
  else
    groups[r] = groups[r] + 1
  end
end
local ret = {}
for i = 1, n do
  local root = uf_findroot(i)
  local cnt = groups[root] - 1
  for dst, _u in pairs(edge[i]) do
    local r = uf_findroot(dst)
    if r == root then cnt = cnt - 1 end
  end
  for dst, _u in pairs(avoid[i]) do
    local r = uf_findroot(dst)
    if r == root then cnt = cnt - 1 end
  end
  ret[i] = cnt
end
print(table.concat(ret, " "))