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 9.2.1) (judge verdict: Runtime Error)

#include <stdio.h>

int main(void){

    int s[10000],a1[20000],a2[20000];
    char c;
    int i,q,t,f,counter=1,sp1=0,sp2=0;
    int n = 0;

    while((c=getchar())!='\n'){
        s[n++] = (int)c;
    }
    rewind(stdin);
    scanf("%d",&q);

    for(i=0;i<q;i++){
        scanf("%d",&t);
        if(t==1){
            counter*=-1;
        }else{
            scanf("%d %c",&f,&c);
            if(f==1){
                if(counter == 1){
                    a1[sp1++] = (int)c;
                }else{
                    a2[sp2++] = (int)c;
                }
            }else{
                if(counter == 1){
                    a2[sp2++] = (int)c;
                }else{
                    a1[sp1++] = (int)c;
                }
            }
            rewind(stdin);
        }
        rewind(stdin);
    }

    if(counter == 1){

        for(i=sp1-1;i>=0;i--){
            printf("%c",(char)a1[i]);
        }

        for(i=0;i<n;i++){
            printf("%c",(char)s[i]);
        }

        for(i=0;i<sp2;i++){
            printf("%c",(char)a2[i]);
        }

    }else{

        for(i=sp2-1;i>=0;i--){
            printf("%c",(char)a2[i]);
        }

        for(i=0;i<n;i++){
            printf("%c",(char)s[i]);
        }

        for(i=0;i<sp1;i++){
            printf("%c",(char)a1[i]);
        }

    }

    return 0;
}