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 1.1.14) (judge verdict: Accepted)

(defun solve (s n queries)
  (declare (ignore n))
  (let ((reversed-p nil))
    (dolist (query (reverse queries))
      (destructuring-bind (x &optional y z) query
        (cond
          ((= x 1) (setf reversed-p (not reversed-p)))
          (t
           (when (or (and reversed-p
                          (= y 2))
                     (and (not reversed-p)
                          (= y 1)))
             (princ z))))))
    (if reversed-p
        (loop for i from (1- (length s)) downto 0
              do (write-char (aref s i)))
        (write-string s))
    (dolist (query queries)
      (destructuring-bind (x &optional y z) query
        (cond
          ((= x 1) (setf reversed-p (not reversed-p)))
          (t
           (unless (or (and reversed-p
                            (= y 2))
                       (and (not reversed-p)
                            (= y 1)))
             (princ z))))))
    (fresh-line)))

#-swank
(let* ((s (read-line))
       (n (read))
       (queries (loop repeat n
                      for x = (read)
                      if (= x 1)
                      collect (list x)
                      else
                      collect (list x (read) (read-char)))))
  (solve s n queries))