Problem:
Score : 300 points

Problem Statement

Given any integer x, Aoki can do the operation below.

Operation: Replace x with the absolute difference of x and K.

You are given the initial value of an integer N. Find the minimum possible value taken by N after Aoki does the operation zero or more times.

Constraints

0 ≤ N ≤ 10^{18}

1 ≤ K ≤ 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the minimum possible value taken by N after Aoki does the operation zero or more times.

Sample Input 1

7 4

Sample Output 1

1

Initially, N=7.

After one operation, N becomes |7-4| = 3.

After two operations, N becomes |3-4| = 1, which is the minimum value taken by N.

Sample Input 2

2 6

Sample Output 2

2

N=2 after zero operations is the minimum.

Sample Input 3

1000000000000000000 1

Sample Output 3

0

Solution in Haskell (GHC 8.8.3) (judge verdict: Accepted)

{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE BangPatterns #-}
{-# LANGUAGE TypeFamilies #-}
import           Data.Maybe
import           Data.Int
import           Data.List
import           Data.Char
  -- import qualified Data.Vector.Unboxing as VU
import qualified Data.Vector as V
import qualified Data.Vector.Mutable as VM
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.ByteString.Char8 as BS
import           Data.List (scanl')
import           Data.IORef
import           Control.Monad
import           Control.Monad.ST
import           Control.Monad.State.Strict
import           Control.Applicative
import           Debug.Trace
import           Text.Printf


readInt = BS.readInt . BS.dropWhile isSpace
getVector f = VU.unfoldr f <$> BS.getLine
getIntVector = getVector readInt
getBSVector = getVector BS.uncons
getInt = readLn :: IO Int
getIntTuple = (\vec -> (vec VU.! 0, vec VU.! 1)) <$> getIntVector

main = do
  (n,k) <- getIntTuple
  print $ solve n k

solve n k = min min1 min2
  where
    times  = n `div` k
    times' = (n + (k-1)) `div` k
    min1 = abs $ (n - times *k)
    min2 = abs $ (n - times'*k)