Problem:
Score : 400 points

Problem Statement

Takahashi has a string S consisting of lowercase English letters.

Starting with this string, he will produce a new one in the procedure given as follows.

The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:

If T_i = 1: reverse the string S.

If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.

If F_i = 1 : Add C_i to the beginning of the string S.

If F_i = 2 : Add C_i to the end of the string S.

Help Takahashi by finding the final string that results from the procedure.

Constraints

1 \leq |S| \leq 10^5

S consists of lowercase English letters.

1 \leq Q \leq 2 \times 10^5

T_i = 1 or 2.

F_i = 1 or 2, if provided.

C_i is a lowercase English letter, if provided.

Input

Input is given from Standard Input in the following format:

S
Q
Query_1
:
Query_Q

In the 3-rd through the (Q+2)-th lines, Query_i is one of the following:

1

which means T_i = 1, and:

2 F_i C_i

which means T_i = 2.

Output

Print the resulting string.

Sample Input 1

a
4
2 1 p
1
2 2 c
1

Sample Output 1

cpa

There will be Q = 4 operations. Initially, S is a.

Operation 1: Add p at the beginning of S. S becomes pa.

Operation 2: Reverse S. S becomes ap.

Operation 3: Add c at the end of S. S becomes apc.

Operation 4: Reverse S. S becomes cpa.

Thus, the resulting string is cpa.

Sample Input 2

a
6
2 2 a
2 1 b
1
2 2 c
1
1

Sample Output 2

aabc

There will be Q = 6 operations. Initially, S is a.

Operation 1: S becomes aa.

Operation 2: S becomes baa.

Operation 3: S becomes aab.

Operation 4: S becomes aabc.

Operation 5: S becomes cbaa.

Operation 6: S becomes aabc.

Thus, the resulting string is aabc.

Sample Input 3

y
1
2 1 x

Sample Output 3

xy

Solution in Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

{-# LANGUAGE OverloadedStrings #-}
{-# 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
  s <- BS.getLine
  q <- getInt
  qq <- V.replicateM q $ getLine
  print $ solve qq s

solve qq s = V.foldl' check s qq
  where
    check s' x
      | length x == 1 = operateT1 s'
      | otherwise = operateT2 s' (x !! 2) (x !! 4)


operateT1 s = if BS.length s == 1
              then s
              else (l `BS.cons` middle) `BS.snoc` h
  where
    h = BS.head s
    l = BS.last s
    middle = BS.tail . BS.init $ s

operateT2 :: BS.ByteString -> Char -> Char -> BS.ByteString
operateT2 s f c  = case f of '1' -> operateF1
                             '2' -> operateF2
                             _ -> undefined
  where
    operateF1 = c `BS.cons` s
    operateF2 = s `BS.snoc` c