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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>

void char_push(char *s, int *len, char inp)
{
   // printf("%d %c\n", *len, inp);
   s[*len] = inp;
   *len = *len + 1;
   s[*len] = '\0';
   return;
}

int main(void)
{
   int Q;
   char S[3 * 100000];
   char S_pre[1 * 100000];
   int T;
   int resF;
   char response;
   int Spre_i=0, Spost_i = 0, arrow = 0;

   scanf("%s", S);
   scanf("%d", &Q);

   for (Spost_i = 0; S[Spost_i] != '\0'; Spost_i++)
   {
   }

   // printf ("%d\n", Spost_i);
   for (int i = 0; i < Q; i++)
   {

      scanf("%d", &T);
      if (T == 1)
      {
         arrow = arrow ^ 1;
      }
      else
      {
         scanf("%d %c", &resF, &response);
         resF = 2 - resF;
         // resF = 0  -> 後ろに
         if ((resF ^ arrow) == 0)
         { char_push(S,&Spost_i, response );}
         else
         { char_push(S_pre, &Spre_i, response);}
         
      }
   }
   if (arrow == 0)
   {
      for (int i = 0; i < Spre_i; i++)
      {
         printf("%c", S_pre[Spre_i -1  - i]);
      }
      printf("%s\n", S);
   }
   else
   {
      for (int i = 0; i < Spost_i; i++)
      {
         printf("%c", S[Spost_i-1 - i]);
      }
      printf("%s\n", S_pre);
   }
   return 0;
}