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: Wrong Answer)

from __future__ import print_function

from functools import reduce
from operator import mul
from collections import Counter
from collections import deque
from itertools import accumulate
from queue import Queue
from queue import PriorityQueue as pq
from heapq import heapreplace
from heapq import heapify
from heapq import heappushpop
from heapq import heappop
from heapq import heappush
import heapq
import time
import random
import bisect
import itertools
import collections
from fractions import Fraction
import fractions
import string
import math
import operator
import functools
import copy
import array
import re
import sys
sys.setrecursionlimit(500000)


input = sys.stdin.readline


def eprint(*args, **kwargs):
    print(*args, file=sys.stderr, **kwargs)
    return

# from fractions import gcd
# from math import gcd

# def lcm(n, m):
#     return int(n * m / gcd(n, m))


# def coprimize(p, q):
#     common = gcd(p, q)
#     return (p // common, q // common)


# def find_gcd(list_l):
#     x = reduce(gcd, list_l)
#     return x


def combinations_count(n, r):
    r = min(r, n - r)
    numer = reduce(mul, range(n, n - r, -1), 1)
    denom = reduce(mul, range(1, r + 1), 1)
    return numer // denom


def main():
    #
    s0 = input().strip()
    q = int(input().strip())

    #
    stc1 = ""
    stc2 = ""
    flag = 0

    # def f(flag1, flag2, val):
    #     if flag1 ^ flag2:
    #         stc1 += val
    #     else:
    #         stc2 += val

    #
    for i in range(q):
        t, *fc = map(lambda x: int(x) if x.isdecimal() else x, input().strip().split())
        # eprint('t,fc ', end=':\n')
        # eprint(t, fc, "\n")

        if t == 2:    # 文字列付加
            # f(fc[0], flag, fc[1])
            if (fc[0]-1)^flag:  # (fc[0]-1)は前方/後方， flagは非反転/反転，をそれぞれ表す
                stc2+=fc[1]
            else:
                stc1+=fc[1]
        else:       # 反転命令
            flag = flag ^ 1

    #
    if flag == 0:
        ans = stc1[::-1] + s0 + stc2
    else:
        ans = stc2[::-1] + s0 + stc1
    print(ans)
    # eprint(ans[::-1])
    return


if __name__ == '__main__':
    main()