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: Accepted)

import qualified Data.ByteString.Char8 as B
import qualified Data.Sequence as SQ

main :: IO ()
main = do
    s <- B.getLine :: IO B.ByteString
    q <- readLn :: IO Int

    let
        solve :: (SQ.Seq Char, SQ.Seq Char) -> Int -> Int -> IO (SQ.Seq Char, SQ.Seq Char, Int)
        solve (sqx,sqy) n cnt
            | n == 0 = return (sqx,sqy,cnt) 
            | otherwise = do
                xs <- words <$> getLine :: IO [String]
                case (length xs) of
                    1 -> do
                        solve (sqx,sqy) (n-1) (cnt+1)
                    _ -> do
                        let [_,f,[c]] = xs
                        case (f, even cnt) of
                            ("1", True) -> solve ((SQ.<|) c sqx, sqy) (n-1) cnt
                            ("1", False) -> solve (sqx, (SQ.|>) sqy c) (n-1) cnt
                            ("2", True) -> solve (sqx, (SQ.|>) sqy c) (n-1) cnt
                            ("2", False) -> solve ((SQ.<|) c sqx, sqy)  (n-1) cnt
        
        outl :: SQ.Seq Char -> IO ()
        outl sq =
            case SQ.viewl sq of
                c SQ.:< sq' -> do
                    putChar c
                    outl sq'
                SQ.EmptyL -> return ()

        outr :: SQ.Seq Char -> IO ()
        outr sq =
            case SQ.viewr sq of
                sq' SQ.:> c -> do
                    putChar c
                    outr sq'
                SQ.EmptyR -> return ()


    (sqx,sqy,cnt) <- solve (SQ.empty, SQ.empty) q 0

    if (even cnt)
        then do
            outl sqx
            B.putStr s
            outl sqy
            putStrLn ""
        else do
            outr sqy
            B.putStr $ B.reverse s
            outr sqx
            putStrLn ""