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 (Clang 3.8.0) (judge verdict: Accepted)

#include <inttypes.h>
#include <math.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void)
{
    char S[(int)1e5 + 2] = {};
    intmax_t Q;
    struct query {
        intmax_t T;
        intmax_t F;
        char C;
    } query[(int)2e5];

    char head[(int)2e5 + 2] = {}, foot[(int)2e5 + 2] = {};
    char *sp, *hp = head + 1, *fp = foot + 1;
    intmax_t reverse = 0;

    scanf("%s", &S[1]);
    scanf("%jd", &Q);
    for (size_t i = 0; i < Q; i++) {
        scanf("%jd", &query[i].T);
        if (query[i].T == 2) {
            scanf("%jd %c", &query[i].F, &query[i].C);
        }
    }

    for (size_t i = 0; i < Q; i++) {
        if (query[i].T == 1) {
            reverse++;
        } else {
            if (query[i].F == 1) {
                if (reverse % 2 == 0) {
                    *hp++ = query[i].C;
                } else {
                    *fp++ = query[i].C;
                }
            } else {
                if (reverse % 2 == 0) {
                    *fp++ = query[i].C;
                } else {
                    *hp++ = query[i].C;
                }
            }
        }
    }

    sp = S + strlen(&S[1]) + 1;
    if (reverse % 2 == 0) {
        while (*--hp) {
            putchar(*hp);
        }
        printf("%s", &S[1]);
        printf("%s", &foot[1]);
    } else {
        while (*--fp) {
            putchar(*fp);
        }
        while (*--sp) {
            putchar(*sp);
        }
        printf("%s", &head[1]);
    }
    printf("\n");

    return 0;
}