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

#define false 0
#define true (!false)
#define STRMAX (300000)

char S[STRMAX];
char pre[STRMAX];
char post[STRMAX];

int revf = false;
int Q;

void revers( char *s )
{
    int i;
    int len = strlen( s );
    char buf[STRMAX];

    strcpy( buf, s );
    for( i=0; i<len; i++ ) s[i] = buf[len - i-1];
}

int main( int argc, char *argv[] )
{
    int i;
    int t, f, c;
    char buf[128];
    char *p1 = pre;
    char *p2 = post;

    scanf( "%s", S );
    scanf( "%d", &Q );

    for( i=0; i<Q; i++ ) {

//printf( "*** %d\n", i );

        scanf( "%d", &t );
//        int n = scanf( "%d%d %c", &t, &f, &c );
//printf( "n = %d\n", n );
        if( t == 1 ) {
            revf = ! revf;
        } else if( t == 2 ) {
            scanf( "%d %c", &f, &c );
            if( f == 1 ) {
                if( ! revf ) *p1 ++ = c;
                else         *p2 ++ = c;
            } else if( f == 2 ) {
                if( ! revf ) *p2 ++ = c;
                else         *p1 ++ = c;
            }
        }
    }

    *p1 = 0;
    *p2 = 0;

    if( ! revf ) {
        revers( pre );
        strcat( pre, S );
        strcpy( S, pre );
        strcat( S, post );
    } else {
        revers( S );
        revers( post );
        strcat( post, S );
        strcpy( S, post );
        strcat( S, pre );
    }

    puts( S );
}