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: Time Limit Exceeded)

#! /usr/bin/env python3

from fractions import gcd
from collections import Counter, deque, defaultdict
from heapq import heappush, heappop, heappushpop, heapify, heapreplace, merge
from bisect import bisect_left, bisect_right, bisect, insort_left, insort_right, insort
from itertools import accumulate, product, permutations, combinations, combinations_with_replacement

S = input()
Q = int(input())

dire = 0
front = []
back = []

for i in range(Q):
    quary = [_ for _ in input().split()]
    if len(quary) == 1:
        T = quary[0]
        dire += 1
    else:
        T, F, C = quary
        if F == "1":
            if dire % 2 == 0:
                front += [C]
            else:
                back += [C]
        else:
            if dire % 2 == 1:
                front += [C]
            else:
                back += [C]

for f in front:
    S = f + S
for b in back:
    S = S + b

if dire % 2 == 1:
    print(S[::-1])
else:
    print(S)