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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <malloc.h>
#include <string.h>
#include <stdlib.h>

//abc158D
int main(void) {

  int s, q, i, j, ti, fi, inv=0, top=0, head=0, tail=0;
  //inv=1:反転、top=0:先頭追加
  char t;
  scanf(" %c %d", &t, &q);
  char str[2 * q + 1];
  for(i=0; i < 2 * q + 1; i++)
    str[i]=0;
  str[q]=t;
  head=q - 1, tail=q + 1;

  for(i=0; i < q; i++){
    scanf(" %d", &ti);

    if(ti == 1) inv=!inv;
    else{
      scanf(" %d", &fi);
      if(fi == 1){
        top=0;
      }else{
        top=1;
      }
      scanf(" %c", &t);
      if(inv + top == 0 || inv + top == 2){
        str[head--]=t;
      }

      else{
        str[tail++]=t;
      }
    }
  }
  if(inv == 0){
    for(i=0; i < 2 * q + 1; i++)
      if(str[i] != 0) printf("%c", str[i]);
  }else{
    for(i=2 * q + 1; i > 0; i--)
      if(str[i] != 0) printf("%c", str[i]);
  }
}