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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>


typedef struct p_tag{
    char c;
    struct p_tag *prev;
    struct p_tag *next;
} string;

int main(){
    char str[100002];
    int qn;
    int i;
    char q[10];
    int rot = 0;
    string *head, *tail, *buf, *buf_prev=NULL;

    (void)fgets(str, 100002, stdin);

    i = 0;
    while(str[i] != '\n'){
        buf = malloc(sizeof(string));
        if(i == 0){
            head = buf;
        }
        buf->c = str[i];
        buf->prev = buf_prev;
        buf->next = NULL;
        if(buf_prev != NULL){
            buf_prev->next = buf;
        }
        buf_prev = buf;
        i++;
    };
    tail = buf;

    (void)scanf("%d", &qn);
    (void)getchar();

    for(i=0; i < qn; i++){
        (void)fgets(q, 10, stdin);
        if(q[0] == '1'){
            rot++;
        }else{
            if(q[2] == '1' && rot % 2 == 0 || q[2] == '2' && rot % 2 == 1){
                buf = malloc(sizeof(string));
                buf->c = q[4];
                buf->next = NULL;
                buf->prev = tail;
                tail->next = buf;
                tail = buf;
            }else{
                buf = malloc(sizeof(string));
                buf->c = q[4];
                buf->next = head;
                buf->prev = NULL;
                head->prev = buf;
                head = buf;
            }
        }
    }
    if(rot % 2 == 1){
        putchar(head->c);
        while(1){
            head = head->next;
            if(head == NULL){
                putchar('\n');
                return 0;
            }
            putchar(head->c);
        }
    }else{
        putchar(tail->c);
        while(1){
            tail = tail->prev;
            if(tail == NULL){
                putchar('\n');
                return 0;
            }
            putchar(tail->c);
        }
    }
}