Problem:
Score : 200 points

Problem Statement

Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the 0 key, the 1 key and the backspace key.

To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string:

The 0 key: a letter 0 will be inserted to the right of the string.

The 1 key: a letter 1 will be inserted to the right of the string.

The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted.

Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter 0 stands for the 0 key, the letter 1 stands for the 1 key and the letter B stands for the backspace key. What string is displayed in the editor now?

Constraints

1 ≦ |s| ≦ 10 (|s| denotes the length of s)

s consists of the letters 0, 1 and B.

The correct answer is not an empty string.

Input

The input is given from Standard Input in the following format:

s

Output

Print the string displayed in the editor in the end.

Sample Input 1

01B0

Sample Output 1

00

Each time the key is pressed, the string in the editor will change as follows: 0, 01, 0, 00.

Sample Input 2

0BB1

Sample Output 2

1

Each time the key is pressed, the string in the editor will change as follows: 0, (empty), (empty), 1.

Solution in Haskell (GHC 7.10.3) (judge verdict: Accepted)

{-# LANGUAGE BangPatterns #-}
 
module Main
  ( main
  )
where
 
import           Control.Applicative
import           Control.Monad
import           Control.Monad.ST
import           Data.Bits
import qualified Data.ByteString.Char8         as BS
import           Data.Char
import           Data.List
import           Data.Maybe
import           Data.Ord
import qualified Data.Vector                   as V
import qualified Data.Vector.Mutable           as MV
import qualified Data.Vector.Unboxed           as UV

parse = fst . fromJust
parseInt = parse . BS.readInt
parseInteger = parse . BS.readInteger
readInt = parseInt <$> BS.getLine
readInteger = parseInteger <$> BS.getLine
readIntList = map parseInt . BS.words <$> BS.getLine
readIntegerList = map parseInteger . BS.words <$> BS.getLine


main = do
  s <- getLine
  putStrLn $ parseStr s ""

parseStr "" stack = reverse stack
parseStr ('0':rest) stack = parseStr rest ('0':stack)
parseStr ('1':rest) stack = parseStr rest ('1':stack)
parseStr ('B':rest) (x:stack) = parseStr rest stack
parseStr ('B':rest) stack = parseStr rest stack