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 8.8.3) (judge verdict: Accepted)

import Control.Monad
import qualified Data.ByteString.Char8 as BS
import Data.Maybe

readInt = fst . fromJust . BS.readInt

readString = map BS.unpack . BS.words

getInt = readInt <$> BS.getLine

getString = readString <$> BS.getLine

getNString n = map readString <$> replicateM n BS.getLine

calc :: String -> [[Char]] -> Int -> String -> String -> String
calc a [] d fs rs'
  | d == 1 = fs ++ a ++ reverse rs'
  | d == (-1) = rs' ++ reverse a ++ reverse fs
calc a (q : qs) d fs rs'
  | q' == '1' = calc a qs (- d) fs rs'
  | d == 1 && f == '1' = calc a qs d (c : fs) rs'
  | d == 1 && f == '2' = calc a qs d fs (c : rs')
  | d == (-1) && f == '1' = calc a qs d fs (c : rs')
  | d == (-1) && f == '2' = calc a qs d (c : fs) rs'
  where
    q' = head q
    f = if q' == '2' then head (tail q) else '0'
    c = if q' == '2' then head (tail (tail q)) else '_'

main = do
  [a] <- getString
  n <- getInt
  qs <- getNString n
  let qs' = map (\q -> map head q) qs
  putStrLn $ calc a qs' 1 "" ""