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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

def msf(s,q,ql):
    len_ql = len(ql)
    for i in range(1,len_ql-1):
        t1 = ql[i][0]
        if t1 == "1":
            s = list(reversed(s))
        elif t1 == "2":
            f = ql[i][1]
            c = list(ql[i][2])
            if f == "1":
                s = c + s
            elif f == "2":
                s = s + c
    return s
    
if __name__ == '__main__':
    s = list(input())
    q = int(input())
    ql = [list(input().split()) for _ in range(q)]

    ql = [['0']] + ql + [['0']]
    len_ql = len(ql)

    for i in range(1,len_ql-1):
        if len_ql <= i:
            break
        elif ql[i][0] == "1" and ql[i-1][0] == "1" :
            ql = ql[0:i-1] + ql[i+1:]
            len_ql -= 2
        else:
            pass

    print(''.join(msf(s,q,ql)))