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>
#define MAXCHAR 100005

char word[MAXCHAR];

void addToBack (char);
void addToFront (char);
void reverse (int, int);

int main ()
{
    char c;
    int q, f, t;
    scanf("%s", word);
    scanf("%d", &q);

    for (int i = 0; i < q; i++)
    {
        scanf("%d", &t);
        if (t == 1)
        {
            int j = 0;
            while (word[j] != '\0')
                j++;
            reverse(0, j - 1);
        }
        else if (t == 2)
        {
            scanf("%d %c", &f, &c);
            getchar();
            if (f == 2)
            {
                addToBack(c);
            }
            else 
            {
                addToFront(c);
            }
        }
    }

    printf("%s\n", word);

    return 0;
}

void addToBack(char c)
{
    int i = 0;
    while (word[i] != '\0')
        i++;
    word[i] = c;
    word[i + 1] = '\0';
}

void addToFront(char c)
{
    int i = 0;
    while (word[i] != '\0')
        i++;
    int len = i;
    while (i >= 0)
    {
        word[i + 1] = word[i];
        i--;
    }
    word[0] = c;
    word[len + 2] = '\0';
}

void swap (char *a, char *b)
{
    char temp;
    temp = *a;
    *a = *b;
    *b = temp;
}

void reverse (int left, int right)
{
    while (right - left > 0)
    {
        swap(&word[left], &word[right]);
        right--;
        left++;
    }
}