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 <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdio.h>
int compare(const void *a,const void *b){
    if(*(int *)a>*(int *)b)return 1;
    return -1;
}
int main()
{   char s[100001];
    scanf("%s",s);
    int q;
    scanf("%d",&q);
    int flag=0,fn=0,rn=0;
    int  fadd[200001]={-1},radd[200001]={-1};
    for(int i=0;i<q;i++){
        int tmp;
        scanf("%d",&tmp);
        if(tmp==1){
            if(flag==1)flag=0;else flag=1;
        }else{
            int f;
            char a[2];
            scanf("%d",&f);
            scanf("%s",a);
            if(f==1&&flag==0){
                fadd[fn]=a[0]-'a';
                fn++;
            }else if(f==1&&flag==1){
                radd[rn]=a[0]-'a';
                rn++;
            }else if(f==2&&flag==0){
                radd[rn]=a[0]-'a';
                rn++;
            }else{
                fadd[fn]=a[0]-'a';
                fn++;
            }
        }
    }
  if(flag==0){
        for(int i=fn-1;i>=0;i--){
            printf("%c",fadd[i]+'a');
        }
        printf("%s",s);
        for(int i=0;i<rn;i++){
            printf("%c",radd[i]+'a');
        }
    }else{
        for(int i=rn-1;i>=0;i--){
        printf("%c",radd[i]+'a');
        }
        int l=strlen(s);
        for(int i=l-1;i>=0;i--){
            printf("%c",s[i]);
        }
        for(int i=0;i<fn;i++){
            printf("%c",fadd[i]+'a');
        }
    }
    

    return 0;
}