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 Haskell (GHC 7.10.3) (judge verdict: Time Limit Exceeded)

import qualified Data.Vector.Unboxed as U
import qualified Data.ByteString.Char8 as C
import Data.List
import Control.Monad
import Control.Arrow
import qualified Data.IntMap as IM
import qualified Data.IntSet as IS
import Data.Maybe
import Data.Graph

r :: Int -> IO (U.Vector Int)
r n = U.unfoldrN n (C.readInt . C.dropWhile (==' ')) <$> C.getLine

r' :: IO [Int]
r'  =   unfoldr    (C.readInt . C.dropWhile (==' ')) <$> C.getLine

main :: IO ()
main = do
    [n,m,k] <- r' :: IO [Int]
    friends <- replicateM m (r' >>= (\[a,b] -> return (a,b)))
    blocks  <- replicateM k (r' >>= (\[s,c] -> return (s,c)))

    let fs = friends ++ map swap friends
    let bs = blocks  ++ map swap blocks

    let fg = buildG (1,n) fs

    let cm = IM.fromList $ map (id &&& (IS.fromList.reachable fg)) [1..n]
    let fm = IM.map IS.fromList $ IM.fromListWith (++) $ map (second pure) fs
    let bm = IM.map IS.fromList $ IM.fromListWith (++) $ map (second pure) bs

    let net = Net n cm fm bm

    putStrLn $ unwords $ map (show .solve net) [1..n]


data Network = Net {population :: Int
    , componentMap :: IM.IntMap IS.IntSet
    , friendMap :: IM.IntMap IS.IntSet
    , blockMap :: IM.IntMap IS.IntSet}

solve net me = IS.size $ componentMap net IM.! me IS.\\
    (IS.fromList [me] `IS.union` (IM.findWithDefault IS.empty me (friendMap net)) `IS.union` (IM.findWithDefault IS.empty me (blockMap net)))

swap (a,b) = (b,a)