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 <stdlib.h>
#include <string.h>
#include <strings.h>

#ifdef TEST
struct test {
    int ans;
} td[] = {
    {},
};
#endif

void
solver(char *s, int q, int *t, int *f, char *c)
{
    int i;
    int top = 0;
    char pre[200001];
    char post[200001];

    bzero(pre, sizeof(pre));
    bzero(post, sizeof(post));

    for(i = 0; i < q; i++) {
        if(t[i] == 1) {
            top = 1 - top;
        }
        if(t[i] == 2) {
            if(f[i] == 1) {
                if(top == 0) {
                    pre[strlen(pre)] = c[i];
                } else {
                    post[strlen(post)] = c[i];
                }
            } else {
                if(top == 1) {
                    pre[strlen(pre)] = c[i];
                } else {
                    post[strlen(post)] = c[i];
                }
            }
        }
    }
    if(top == 0) {
        for(i = strlen(pre) - 1; i >= 0; i--)
            printf("%c", pre[i]);
        for(i = 0; i < strlen(s); i++)
            printf("%c", s[i]);
        for(i = 0; i < strlen(post); i++)
            printf("%c", post[i]);
    } else {
        for(i = strlen(post) - 1; i >= 0; i--)
            printf("%c", post[i]);
        for(i = strlen(s) - 1; i >= 0; i--)
            printf("%c", s[i]);
        for(i = 0; i < strlen(pre); i++)
            printf("%c", pre[i]);
    }
    printf("\n");
}

int
main(int argc, char *argv[])
{
    char S[100001];
    char tmp[100];
    int Q;
    int T[200000];
    int F[200000];
    char C[200000];
    int i;

#ifdef TEST
    int tt;
    for(tt = 0; tt < sizeof(td) / sizeof(struct test); tt++) {
        solver();
        printf("%d\n", td[tt].ans);
    }
#endif
    scanf("%s", S);
    scanf("%d", &Q);
    for(i = 0; i < Q; i++) {
        scanf("%d", T + i);
        if(T[i] == 2) {
            scanf("%d %c", F + i, C + i);
        }
    }

    solver(S, Q, T, F, C);
    return 0;
}