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 Common Lisp (SBCL 2.0.3) (judge verdict: Accepted)

(let* ((s (read-line))
       (q (read))
       (head nil)
       (tail nil)
       (rev 0))
  (loop :for i :from 1 :to q
        :for x := (read)
        :if (= x 1)
        :do (progn
              (incf rev))
        :else
        :do (let ((f (read))
                  (c (read-char)))
              (cond ((and (evenp rev) (= f 1))
                     (push c head))
                    ((and (evenp rev) (= f 2))
                     (push c tail))
                    ((and (oddp rev) (= f 1))
                     (push c tail))
                    (t
                     (push c head)))))
  (when (oddp rev)
    (rotatef head tail)
    (setf s (reverse s)))
  (loop :for c :in head :do (format t "~A" c))
  (format t "~A" s)
  (loop :for c :in (reverse tail) :do (format t "~A" c))
  (format t "~%"))