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 Control.Monad
import Control.Monad.ST
import Data.Array.ST
import qualified Data.ByteString.Char8 as C
import Data.IntMap.Strict (fromList, (!))
import Data.List
import qualified Data.Set as S

main = sub <$> get >>= put

get = fmap (unfoldr (C.readInt . C.dropWhile (==' '))) . C.lines <$> C.getContents

put = putStrLn . unwords . fmap show

sub ([n,m,k]:as) = sol n (splitAt m as)

sol n (es,bs) = fmap f [1..n]
  where
  rs = runST $ do
    uf <- newUF n
    forM_ es (unionUF uf) 
    rs <- mapM (findUF uf) [1..n]
    return $ fromList $ zip [1..] rs
  se = S.fromList $ pair <$> es
  sb = S.fromList $ pair <$> bs
  f i = length [1 | j <- [1..n], j/=i, rs!i==rs!j, 
    (pair [i,j]) `S.notMember` se, (pair [i,j]) `S.notMember` sb]

pair [a,b] = (min a b, max a b)

-- | Union-Find
data UnionFind s = UnionFind {ids:: STUArray s Int Int, szs:: STUArray s Int Int}

newUF :: Int -> ST s (UnionFind s)
newUF n = liftM2 UnionFind (newListArray (1,n) [1..n]) (newArray (1,n) 1)

findUF :: UnionFind s -> Int -> ST s Int
findUF uf i = do
  ix <- readArray (ids uf) i
  if ix == i
    then return i
    else do
      writeArray (ids uf) i =<< readArray (ids uf) ix
      findUF uf ix

unionUF :: UnionFind s -> [Int] -> ST s ()
unionUF uf [p,q] = do
  i <- findUF uf p
  j <- findUF uf q
  szi <- readArray (szs uf) i
  szj <- readArray (szs uf) j
  if szi < szj
    then do
      writeArray (ids uf) i j
      writeArray (szs uf) j (szi+szj)
      writeArray (szs uf) i 0
    else do
      writeArray (ids uf) j i
      writeArray (szs uf) i (szj+szi)
      writeArray (szs uf) j 0