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 8.8.3) (judge verdict: Accepted)

import Control.Monad
import Data.Maybe
import qualified Data.ByteString.Char8 as BS
import Data.List
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM

newUF n = VUM.replicate n (-1 :: Int)

root uf x = do
    p <- VUM.read uf x
    if p < 0
        then return x
        else do
            r <- root uf p
            VUM.write uf x r
            return r

same uf x y = liftM2 (==) (root uf x) (root uf y)

unite uf x y = do
    px <- root uf x
    py <- root uf y
    when (px /= py) $ do
        sx <- VUM.read uf px
        sy <- VUM.read uf py
        let (par, chld) = if sx < sy then (px, py)　else (py, px)
        VUM.write uf chld par
        VUM.write uf par (sx + sy) 

size uf x = do
    px <- root uf x
    s <- VUM.read uf px
    return $ abs s

readInt = fst . fromJust . BS.readInt
readIntList = map readInt . BS.words
getInt = readInt <$> BS.getLine
getIntList = readIntList <$> BS.getLine

main = do
    [n,m,k] <- getIntList
    ab <- replicateM m $ do
        [a,b] <- getIntList
        return (a-1,b-1)
    cd <- replicateM k $ do
        [c,d] <- getIntList
        return (c-1,d-1)
    let friends = VU.create $ do
            count <- VUM.replicate n (-1 :: Int)
            uf <- newUF n
            forM_ ab $ \(a,b) -> do
                unite uf a b
                VUM.modify count pred a
                VUM.modify count pred b
            forM_ [0..n-1] $ \i -> do
                s <- size uf i
                VUM.modify count (+ s) i
            forM_ cd $ \(c,d) -> do
                b <- same uf c d
                when b $ do
                    VUM.modify count pred c
                    VUM.modify count pred d
            return count
    putStrLn . unwords . map show $ VU.toList friends