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 Python (3.4.3) (judge verdict: Accepted)

from collections import deque
#双方向リストにするため、dequeを使用する
#（先頭と末尾に文字を追加するため）
s = deque(input())
q = int(input())

#一番最後に1度だけ反転すればよい
reverse_count = 0
for i in range(q):
  y = input().split()
  if y[0] == '1':
    #反転は最後に行う
    reverse_count += 1
    continue
  elif y[0] == '2':
    if y[1] == '1':
      #文字列の先頭に追加。反転回数が奇数なら、末尾に追加と同義
      if reverse_count % 2 == 1:
        s.append(y[2])
      else:
        s.appendleft(y[2])
    elif y[1] == '2':
      #文字列の末尾に追加。反転回数が奇数なら、先頭に追加と同義
      if reverse_count % 2 == 1:
        s.appendleft(y[2])
      else:
        s.append(y[2])

#最後に一度だけ反転させる
if reverse_count % 2 == 0:
  #元の順序に戻るので、そのまま
  print("".join(s))
else:
  s.reverse()
  print("".join(s))