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: Accepted)

#include <stdio.h>
#include <string.h>

int main(void) {
    int Q, T, F, l, r, a, s, i;
    char S[100005], L[200005], R[200005], C;
    scanf("%s%d", S, &Q);
    a=0; l=0; r=0;
    for (i=0; i<Q; i++) {
        scanf("%d", &T);
        if (T==1) {
            a++;
            a%=2;
        }
        else {
            scanf("%d %c", &F, &C);
            if ((a+F)%2==1) {
                L[l]=C;
                l++;
            }
            else {
                R[r]=C;
                r++;
            }
        }
    }
    if (a==0) {
        for (i=0; i<l; i++) printf("%c", L[l-i-1]);
        printf("%s%s", S, R);
    }
    else {
        for (i=0; i<r; i++) {
            printf("%c", R[r-i-1]);
        }
        s=strlen(S);
        for (i=0; i<s; i++) {
            printf("%c", S[s-i-1]);
        }
        printf("%s", L);
    }
    
    /*
    printf("\n");
    printf("%s\n%s\n%s\n", L, S, R);
    printf("%d, %d, %d", a, l, r);
    */
    return 0;
}