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: Time Limit Exceeded)

#include<stdio.h>
#include<string.h>
void star(char a[100010], char b)
{
    int i;
    for (i = strlen(a); i > 0; i--)
    {
        a[i] = a[i - 1];
    }
    a[0] = b;
}
int main()
{
    char a[1000100], b;
    int n, m, i, j,k=1,num=0;
    scanf("%s", a);
    scanf("%d", &n);
    for (i = 0; i < n; i++)
    {
        scanf("%d", &m);
        if (m == 1)
        {
            k = k * -1;
            num++;
        }
        if (m == 2)
        {
            scanf("%d %c", &j,&b);
            if (j == 2)
            {
                j = -1;
            }
            if (j*k==1)
            {
                star(a, b);
            }
            if (j*k==-1)
            {
                a[strlen(a)] = b;
            }
        }
    }
    if (num % 2 == 0)
    {
        printf("%s", a);
    }
    else
    {
        for (i = strlen(a)-1; i >= 0; i--)
        {
            printf("%c", a[i]);
        }
    }
    return 0;
}