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>

int main (void) {
    int     i;
    char    s[200001];
    int     q;
    int     t, f;
    char    c;
    char    prefix_[200001], suffix_[200001];
    char    *prefix = prefix_, *suffix = suffix_;
    int     order = 1;

    prefix[0] = '\0';
    suffix[0] = '\0';
    scanf("%s%d", s, &q);

    for ( i=0; i<q; i++ ) {
        scanf("%d", &t);
        if ( t == 1 ) {
            order *= -1;
            char *tmp = prefix;
            prefix = suffix;
            suffix = tmp;
        }
        else {
            scanf("%d %c", &f, &c);
            if ( f == 1 ) {
                int prefix_len = strlen(prefix);
                prefix[prefix_len] = c;
                prefix[prefix_len+1] = '\0';
            }
            else {
                int suffix_len = strlen(suffix);
                suffix[suffix_len] = c;
                suffix[suffix_len+1] = '\0';
            }
        }
    }

    char prefix_r[200001];
    for ( i=strlen(prefix)-1; i>=0; i-- )
        prefix_r[strlen(prefix)-1-i] = prefix[i];
    prefix_r[strlen(prefix)] = '\0';
    printf("%s", prefix_r);
    if ( order == 1 ) {
        printf("%s", s);
    }
    else {
        char s_[200001];
        for ( i=strlen(s)-1; i>=0; i-- )
            s_[strlen(s)-1-i] = s[i];
        s_[strlen(s)] = '\0';
        printf("%s", s_);
    }
    printf("%s\n", suffix);
    return EXIT_SUCCESS;
}