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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int main(void){
 char s[100001];
 char f[1000000];
 char b[1000000];
 int flength=0,blength=0,slength=0;
 int q;
 int q1,q2,state=0,i;
 char q3;
 scanf("%s",s);
 scanf("%d",&q);
 for(i=0;i<q;i++){
   scanf("%d",&q1);
   if(q1==1){
     if(state){
       state=0;
     }else{
       state=1;
     }
   }else{
     q2=' ';
     q3=' ';
     while(q2==' '||q2=='\n')
      scanf("%d",&q2);
     while(q3==' '||q3=='\n')
      scanf("%c",&q3);
     q2+=state;
     if(q2%2==0){
       b[blength++]=q3;
     }else{
       f[flength++]=q3;
     }
   }
 }
 //printf("state=%d\n",state);
 if(state){
   for(i=blength-1;i>=0;i--){
     printf("%c",b[i]);
   }
   for(i=0;i<100001;i++){
     if(s[i]==0){
       slength=i;
       break;
     }
   }
   for(i=slength-1;i>=0;i--){
     printf("%c",s[i]);
   }
   printf("%s\n",f);
 }else{
   for(i=flength-1;i>=0;i--){
     printf("%c",f[i]);
   }
    printf("%s%s\n",s,b);
 }
 return 0;
}