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)

# coding:utf-8

import sys
import math
import time
#import numpy as np
import collections
from collections import deque
import queue
import copy
import bisect


def dfs(s, prev, pre):
  val[s] += Ope[s] + prev
  for i in G[s]:
    if(i == pre):
      continue
    dfs(i, val[s], s)

def bfs(sx, sy, n):
  que = deque([[sy,sx]])
  while que:
    y,x = que.popleft()
    for i,j in D:
      if(x+i<0 or y+j<0 or x+i>W-1 or y+j>H-1):
        continue
      dist = G[y+j][x+i]
      if(G[y+j][x+i] != "X"):
        if(type(G[y+j][x+i]) is str):
          G[y+j][x+i] = G[y][x]+1
          que.append([y+j,x+i])
        elif(dist>G[y][x]+1):
          G[y+j][x+i] = G[y][x]+1
          que.append([y+j,x+i])
        

#X = str(input()).split()
#a = [int(x) for x in input().split()]
#sys.setrecursionlimit(10**7)
#N, Q = map(int, input().split())
#G = [list(input()) for i in range(H)]

S = str(input())
Q = int(input())
flag = 0

S2 = deque(S)

#print(S2)

for i in range(Q):
  Que = str(input()).split()
  if(int(Que[0])==1):
    flag = 1 - flag
  else:
    if(flag == 1):
      if(int(Que[1])==1):
        S2.append(Que[2])
      else:
        S2.appendleft(Que[2])
    else:
      if(int(Que[1])==1):
        S2.appendleft(Que[2])
      else:
        S2.append(Que[2])

S = "".join(list(S2))

if(flag == 1):
  S = S[::-1]

print(S)