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>
#include<stdlib.h>
#include<string.h>
#include<math.h>

#define num sizeof(a)/sizeof(int)
#define N 100000

int main(){
    char s[3*N],add[2],sagyou[3*N],emp[3*N];
    int q,size,t,f,number=0,i,ins,distin=0,repa=0;
    scanf("%s%d",s,&q);
    s[N-1]='\0';
    strcpy(sagyou,s);
    for(size=0;s[size]!='\0';size++);

    while(q!=0){
        scanf("%d",&t);
        if(t==1){
            //hanntenn
            distin+=1;
            distin%=2;
        }else if(t==2){
            scanf("%d",&f);
            if((f==1 && distin%2==0) || (f==2 && distin%2)==1){
                //senntoutuika
                scanf("%s",add);                
                strcat(emp,add);
            }else if((f==2 && distin%2==0) || (f==1 && distin%2==1)){
                //matubituika
                scanf("%s",add);
                strcat(sagyou,add);
                size++;
            }
        }
        q--;
    }

    if(distin%2==0){
        for(i=number+1;i>=0;i--){
            putchar(emp[i]);
        }
        printf("%s\n",sagyou);
    }else{
        for(i=size-1;i>=0;i--){
            putchar(sagyou[i]);
        }
        printf("%s\n",emp);
    }
    return 0;
}