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>
#include <stdlib.h>

void turn(char *str, long long int len) {
    for (long long int i = 0; i < len / 2; i++) {
        char tmp = str[i];
        str[i] = str[len - i - 1];
        str[len - i - 1] = tmp;
    }
    // printf("%s\n", str);
}

int main(void) {
    long long int buf = 1;
    for (int i = 0; i < 5; i++) {
        buf *= 10;
    }
    buf *= 3;
    char str[buf + 1];
    memset(str, '\0', buf + 1);
    long long int queries;

    scanf("%s", str);
    scanf(" %lld", &queries);
    for (long long int i = 0; i < queries; i++) {
        int t;
        scanf(" %d", &t);
        if (t == 1) {
            turn(str, strlen(str));
        } else if (t == 2) {
            int f;
            char c;
            scanf(" %d %c", &f, &c);
            if (f == 1) {
                char tmp[buf + 1];
                strncpy(tmp, str, strlen(str));
                str[0] = c;
                strncpy((str + 1), tmp, strlen(tmp));
            } else if (f == 2) {
                str[strlen(str)] = c;
                str[strlen(str)] = '\0';
            }
        }
        // printf("%s\n", str);
    }

    printf("%s\n", str);
    return 0;
}