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: Accepted)

import math, string, itertools, fractions, heapq, collections, re, array, bisect, copy, functools, random
import sys
from collections import deque, defaultdict, Counter; from heapq import heappush, heappop
from itertools import permutations, combinations, product, accumulate, groupby
from bisect import bisect_left, bisect_right, insort_left, insort_right
from operator import itemgetter as ig
sys.setrecursionlimit(10 ** 7)
inf = 10 ** 20; INF = float("INF"); ans = 0; tmp = 0; cnt = 0; ansli = []; tmpli = []; candili = []; stillset = set()
eps = 1.0 / 10 ** 10; mod = 10 ** 9 + 7
dd = [(-1, 0), (0, 1), (1, 0), (0, -1)]; ddn = dd + [(-1, 1), (1, 1), (1, -1), (-1, -1)]; ddn9 = ddn + [(0, 0)]
"""for dx, dy in dd:
        nx = j + dx; ny = i + dy
            if 0 <= nx < w and 0 <= ny < h:"""
def wi(): return list(map(int, sys.stdin.readline().split()))
def wip(): return [int(x) - 1 for x in sys.stdin.readline().split()]#WideIntPoint
def ws(): return sys.stdin.readline().split()
def i(): return int(sys.stdin.readline())
def s(): return input()
def s_list(): return list(input())
def hi(n): return [i() for _ in range(n)]
def hs(n): return [s() for _ in range(n)]#HeightString
def mi(n): return [wi() for _ in range(n)]#MatrixInt
def num__grid(n): return [[int(i) for i in sys.stdin.readline().split()[0]] for _ in range(n)]#NumberGrid
def mip(n): return [wip() for _ in range(n)]
def ms(n): return [ws() for _ in range(n)]
def grid(n): return [s_list() for _ in range(n)]
# ''.join(s)

s = list(s())
s_deque = deque(s)
q = i()
query = ms(q)

cnt = 0
v = ["r", "l"]
vector = v[cnt]

for qu in query:
    if qu[0] == "1":
        cnt += 1
        vector = v[cnt % 2]
    else:
        if qu[1] == "1":
            if vector == "r":
                s_deque.appendleft(qu[2])
            else:
                s_deque.append(qu[2])
        elif qu[1] == "2":
            if vector == "r":
                s_deque.append(qu[2])
            else:
                s_deque.appendleft(qu[2])

print(''.join(s_deque)[::(-1) ** (cnt % 2)])