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>

int main(void)
{

    char c_inputCharacterBuffer, c_decoy;
    char inputString[100000] = {'\0'}, addForward[100000] = {'\0'}, addBackward[100000] = {'\0'};
    int operateCount, stringCount = 0, addForwardCount = 0, addBackwardCount = 0, operateFlag, reverseFlag = 0, addFlag;
    scanf("%s", inputString);
    for (int index = 0; inputString[index] != '\0'; index++)
    {
        stringCount++;
    }
    scanf("%d", &operateCount);

    for (int index = 0; index < operateCount; index++)
    {
        scanf("%d", &operateFlag);
        if (operateFlag == 1)
        {
            if (reverseFlag == 0)
            {
                reverseFlag = 1;
            }
            else
            {
                reverseFlag = 0;
            }
        }
        else if (operateFlag == 2)
        {
            scanf("%d%c%c", &addFlag, &c_decoy, &c_inputCharacterBuffer);
            if ((reverseFlag == 0 && addFlag == 1) || (reverseFlag == 1 && addFlag == 2))
            {
                addForward[addForwardCount] = c_inputCharacterBuffer;
                addForwardCount++;
            }
            else if ((reverseFlag == 0 && addFlag == 2) || (reverseFlag == 1 && addFlag == 1))
            {
                addBackward[addBackwardCount] = c_inputCharacterBuffer;
                addBackwardCount++;
            }
        }
    }

    if (reverseFlag == 0)
    {
        if (addForwardCount > 0)
        {
            for (int index = addForwardCount - 1; index > -1; index--)
            {
                printf("%c", addForward[index]);
            }
        }
        printf("%s", inputString);
        printf("%s", addBackward);
    }
    else
    {
        if (addBackwardCount > 0)
        {
            for (int index = addBackwardCount - 1; index > -1; index--)
            {
                printf("%c", addBackward[index]);
            }
        }
        for (int index = stringCount - 1; index > -1; index--)
        {
            printf("%c", inputString[index]);
        }
        printf("%s", addForward);
    }

    return 0;
}