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: Wrong Answer)

from collections import deque
d=deque()
S=input()
Q=int(input())
query=[list(map(str,input().split())) for _ in range(Q)]
flag=0  #反転なし
#print(query)
s=S
s2=""
d.append(S)
"""
for i in range(Q):
    #print(i,len(d))
    if (len(query[i])==1):
        if len(d)>1:
            flag=(flag+1)%2 #反転フラグをひっくり返す
    else:
        if flag==0:
            if query[i][1]=="1":
                d.appendleft(query[i][2])
            else:
                d.append(query[i][2])
        else:
            if query[i][1]=="2":
                d.appendleft(query[i][2])
            else:
                d.append(query[i][2])
    #print(d)
if flag==0:
    print("".join(list(d)))
else:
    d.reverse()
    print("".join(list(d)))
"""
for i in range(Q):
    #print(i,len(d))
    if (len(query[i])==1):
        d.reverse()
    else:
        if query[i][1]=="1":
            d.appendleft(query[i][2])
        else:
            d.append(query[i][2])
    #print(d)
print("".join(list(d)))