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



Sample input:
a
4
2 1 p
1
2 2 c
1


Sample output:
cpa

Submission status: Wrong Answer

Solution code:
S =input()

input_line = int(input())

flag0 = ""
flag1 = ""

for i in range(input_line):
    action = list(input())
    
    flag = 0
    

    # print(flag0,flag1)
    # print(action)
    if action[0]=="1":
        if flag == 0:
            flag = 1
        else:
            flag = 0
    else:
        # print("check")
        if action[2] == "2":
            
            if flag == 0:
                flag1 = flag1 + action[4]
            elif flag ==1:
                flag0 = action[4]+flag0
        else:
            if flag == 1:
                flag1 += action[4]
            
            elif flag == 0:
                flag0 = action[4]+flag0

if flag == 1:
    S = S[::-1]
print(flag1)
S = flag1 + S + flag0
    
print(S)

Submission status: Wrong Answer

Solution code:
from collections import deque
s=deque(input())
f=False
for _ in range(int(input())):
  t=list(input().split())
  if t[0]=='1':
    if f is False:
      f = True
  else:
    if t[1]=='1':
      if f is False:
      	s.appendleft(t[2])
      else:
        s.append(t[2])
    else:
      if f is False:
      	s.append(t[2])
      else:
        s.appendleft(t[2])
print(''.join(s))

Submission status: Wrong Answer

Solution code:
S = input()
Q = int(input())
fg = False
que2 = []
que2.append(S)

for _ in range(Q):
    que = list(map(str, input().split()))
    if que[0] == "1":
        fg = not fg
    elif (que[1] == "1" and fg == False) or (que[1] == "2" and fg == True) :
        que2.insert(0,que[2])
    else:
        que2.append(que[2])
        
if fg == True:
  que2.reverse()
 
ans = "".join(que2)
print(ans)

Submission status: Wrong Answer

Solution code:
S = input()
Q = int(input())
fg = False
que2 = []
que2.append(S)

for _ in range(Q):
    que = list(map(str, input().split()))
    if que[0] == "1":
        fg = not fg
    elif (que[1] == "1" and fg == False) or (que[1] == "2" and fg == True) :
        que2.insert(0,que[2])
    else:
        que2.append(que[2])
        
if fg == True:
  que2.reverse()

Submission status: Wrong Answer

Solution code:
S = input()
Q = int(input())
fg = False
que2 = []
que2.append(S)

for _ in range(Q):
    que = list(map(str, input().split()))
    if que[0] == "1":
        que2.reverse()
    elif que[0] == "2" and que[1] == "1":
        que2.insert(0,que[2])
    elif que[1] == "2":
        que2.append(que[2])
        
ans = "".join(que2)
print(ans)

Submission status: Wrong Answer

Solution code:
from collections import deque

S = input()
Q = int(input())
q = deque(S)
fg = False
for i in range(Q):
    que = list(map(str , input().split()))
    if que[0] == "1":
        fg = True
    elif (que[1] == "1" and fg == False) or (que[1] == "2" and fg == True) :
        q.appendleft(que[2])
    else:
        q.append(que[2])
if fg:
    q.reverse()
print("".join(q))

Submission status: Wrong Answer

Solution code:
S = [input()]
Q = int(input())

left = ['']
right = ['']
is_reversed = False

for i in range(Q) :
    query = input().split()
    action_type = query[0]
    if (action_type == "1") :
        is_reversed = not is_reversed
        continue

    insert_type = query[1]
    if (insert_type == "1") :
        if (not is_reversed) :
            # 先頭に追加
            # S.insert(0, query[2])
            left.append(query[2])
        else :
            # S.append(query[2])
            right.append(query[2])
    else :
        if (not is_reversed) :
            # S.append(query[2])
            right.append(query[2])
        else :
            # S.insert(0, query[2])
            left.append(query[2])

# if (not is_reversed) :
#     print(''.join(S))
# else :
#     print(''.join(S[::-1]))


if (not is_reversed) :
    l = ''.join(left[::-1])
    r = ''.join(right)
    s = S[0]
    print( l + s + r )
else :
    l = ''.join(left)
    r = ''.join(right[::-1])
    s = S[0]
    print( r + s + l )

Submission status: Wrong Answer

Solution code:
from collections import deque

s = input()
c = int(input())

que = deque(s)

flipped = False
for i in range(c):
    query = input()
    
    if query[0] == '1':
        flipped = not flipped
    
    else:
        t, f, c = query.split()
        if (f == '1') ^ flipped:
            que.appendleft(c)
        
        else:
            que.append(c)

ans = ''.join(que)

print(ans)

Submission status: Wrong Answer

Solution code:
s = str(input())
n = int(input())

stat = 0
front = ""
back = ""

for _ in range(n):
    q = input()

    # 反転の場合
    if q[0] == "1":
        stat = (stat + 1) % 2
        continue

    # q = q.split()

    judge = int(q[2])
    if stat == 1:
        if judge == 1:
            judge = 2
        elif judge == 2:
            judge = 1

    if judge == 1:
        front = q[4] + front 

    if judge == 2:
        back = back + q[4]

s = front + s + back

print(s)

Submission status: Wrong Answer

Solution code:
s = str(input())
n = int(input())

for _ in range(n):
    q = input()

    # 反転の場合
    if q[0] == "1":
        s = s[::-1]
        continue

    q = q.split()

    print(q)

    if q[1] == "1":
        s = q[2] + s

    if q[1] == "2":
        s = s + q[2]

print(s)

Submission status: Wrong Answer

Solution code:
#!/usr/bin/env python3 ## -*- python -*-
from collections import deque

def main():
    s = deque()
    s.append(input())
    q = int(input())
    head = "l"
    for i in range(q):
        d = input()
        if d[0] == "1":
            if head == "l":
                head = "r"
            else:
                head = "l"
        else: # 2
            dd = d.split()
            f = int(dd[1])
            c = dd[2]
            if f == 1:
                if head == "l":
                    s.appendleft(c)
                else:
                    s.append(c)
            else: # f == 2
                if head == "l":
                    s.append(c)
                else:
                    s.appendleft(c)

    if head == "r":
        s.reverse()
    ans = ''.join(s)
    print(ans)
main()

Submission status: Wrong Answer

Solution code:
S = [input()]
Q = int(input())

is_reversed = False

for i in range(Q) :
    query = input().split()
    action_type = query[0]
    if (action_type == "1") :
        is_reversed = True
        continue

    insert_type = query[1]
    if (insert_type == "1") :
        if (not is_reversed) :
            # 先頭に追加
            S.insert(0, query[2])
        else :
            S.append(query[2])
    else :
        if (not is_reversed) :
            S.append(query[2])
        else :
            S.insert(0, query[2])

if (not is_reversed) :
    print(''.join(S))
else :
    print(''.join(S[::-1]))

Submission status: Wrong Answer

Solution code:
s = list(str(input()))
q = int(input())
for i in range (q):
    t = list(str(input()))
    if t[0] == '2':
        if t[2] == '2':
            s.reverse()
            s.append(t[4])
            s.reverse()      
        else:
            s.append(t[4])
    else:
        s.reverse()
print(''.join(s))

Submission status: Wrong Answer

Solution code:
s = list(str(input()))
q = int(input())
for i in range (q):
    t = list(str(input()))
    if t[0] == '1':
        s.reverse()
    elif t[0] == '2':
        if t[2] == '1':
            s.append(t[4])
        elif t[2] == '2':
            s.reverse()
            s.append(t[4])
            s.reverse()
s.reverse()
print(''.join(s))

Submission status: Wrong Answer

Solution code:
from collections import deque

s = list(str(input()))
q = int(input())
flag = True  # trueの時は反転が偶数回
deq = deque(s)
info = []
for i in range(q):
    info.append(list(input().split()))

for j in info:
    if j[0] == '1':
        if flag:
            flag = False
        else:
            flag = True
    else:
        if j[1] == '1' and flag:
            deq.appendleft(j[2])
        elif j[1] == '1' and (not flag):
            deq.append(j[2])
        elif j[1] == '2' and flag:
            deq.append(j[2])
        else:
            deq.appendleft(j[2])

if info[-1][0] == '1' and (not flag):
    deq = reversed(deq)
deq = "".join(deq)
print(deq)

Submission status: Wrong Answer

Solution code:
from collections import deque
S=deque([input()])
q=int(input())
rev=False
for i in range(q):
   s=input()
   if s=="1":
      rev= not rev
   else:
      t,f,c=s.split()
      if (rev and f=="1") or (not rev and f=="2"):
         S.append(c)
      else:
         S.appendleft(c)
if rev:
   S.reverse()
print("".join(S))

Submission status: Wrong Answer

Solution code:
from collections import deque
S=deque([input()])
q=int(input())
rev=False
for i in range(q):
   s=input()
   if s[0]=="1":
      rev= not rev
   else:
      t,f,c=s.split()
      if (rev and f=="1") or (not rev and f=="2"):
         S.append(c)
      else:
         S.appendleft(c)
if not rev:
   print("".join(S))
else:
   print("".join(reversed(S)))

Submission status: Wrong Answer

Solution code:
from collections import deque
S=deque([input()])
q=int(input())
rev=False
for i in range(q):
   s=input()
   if s=="1":
      rev= not rev
   else:
      t,f,c=s.split()
      if (rev and f=="1") or (not rev and f=="2"):
         S.append(c)
      else:
         S.appendleft(c)
if not rev:
   print("".join(S))
else:
   print("".join(reversed(S)))

Submission status: Wrong Answer

Solution code:
from collections import deque
S=deque([input()])
q=int(input())
rev=False
for i in range(q):
   s=input()
   if s=="1":
      rev= not rev
   else:
      t,f,c=s.split()
      if (rev==False and f=="1") or (rev==True and f=="2"):
         S.appendleft(c)
      else:
         S.append(c)
if rev==False:
   print("".join(S))
else:
   print("".join(reversed(S)))

Submission status: Wrong Answer

Solution code:
S = input()
Q = int(input())

cnt_reverse = 0

for q in range(Q):
  _ = input()
  if _[0] == '1':
    cnt_reverse += 1
  else:
    _T, _F, _C = _.split()
    ## リバース回数が奇数偶数のいずれかをチェック
    if _F == '1':
      if cnt_reverse % 2 == 0:
        S = _C + S
      else:
        S = S + _C
    else:
      if cnt_reverse % 2 == 0:
        S = S + _C
      else:
        S = _C + S

# reverseの部分を対応
if cnt_reverse % 2 == 1:
  S = ''.join(sorted(S))

print(S)

Submission status: Wrong Answer

Solution code:
import sys
sys.setrecursionlimit(1000000)
# input = sys.stdin.readline
from bisect import *
from collections import *
from heapq import *
INF = 500000
mod = 10**9+7

S = input()
Q = int(input())
flg = 0
for i in range(Q):
    query = input().split()
    if query[0] == '1':
        flg ^= 1
    else:
        F, C = int(query[1])-1, query[2]
        if F^flg:
            S = C+S
        else:
            S += C
if flg:
    print(S)
else:
    print(S[::-1])

Submission status: Wrong Answer

Solution code:
from collections import deque

s = input()
q = int(input())
word = deque()
rev = 0 # revが0のとき先頭、1のとき後尾
word.append(s)

for _ in range(q):
    t = input().split()

    if len(t) == 1:
        rev ^= 1
    else:
        if rev == 1:
            if t[1] == '1':
                word.append(t[2])
            else:
                word.appendleft(t[2])
        else:
            if t[1] == '1':
                word.appendleft(t[2])
            else:
                word.append(t[2])

if rev == 1:
    word.reverse()

for i in range(len(word)):
    print(word[i],end='')

print()

Submission status: Wrong Answer

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):   
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array:
            # ex) [[0] * N for _ in range(N)]
            # ex2) dp = [[0] * (N+1) for _ in range(W*2)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    r_float = lambda: float(r())
    R = lambda: list(map(int, r().split()))
    R_str = lambda: r().split()
    R_map = lambda: map(int, r().split())
    R_float = lambda: list(map(float, r().split()))
    R_tuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main
    
    # l1 = []
    # l2 = []
    l1 = [0] * (10**6)
    index_1 = 0
    l2 = [0] * (10**6)
    index_2 = 0
    status = 0
    
    def update(status):
        if status == 0:
            return 1
        else:
            return 0
    
    s = r()
    q = r_int()
    for _ in range(q):
        qq = r()
        if qq[0] == '1':
            status = update(status)
        else:
            _, b, c = qq.split()
            if (int(b) + status)%2 == 0:
                # l2.append(c)
                l2[index_2] = c
                index_2 += 1
            else:
                # l1.append(c)
                l1[index_1] = c
                index_1 += 1
    if status == 0:
        # print("".join(reversed(l1)), end="")
        # print("".join(s), end="")
        # print("".join(l2))
        
        # print("".join(reversed(l1[:index_1])), end="")
        # print("".join(s), end="")
        # print("".join(l2[:index_2]))
        
        print(*l1[index_1-1::-1], sep="", end="")
        print(s, sep="", end="")
        print(*l2[:index_2], sep="", end="")
        print()
    else:
        # print("".join(reversed(l2)), end="")
        # print("".join(s[::-1]), end="")
        # print("".join(l1))
        print("".join(reversed(l2[:index_2])), end="")
        print("".join(s[::-1]), end="")
        print("".join(l1[:index_1]))
    # to finish code, use return instead of exit()
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()