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 Go (1.6) (judge verdict: Time Limit Exceeded)

package main
     
import (
    "fmt"
)
 
func main() {
    var s string
    var q int
    fmt.Scan(&s)
    fmt.Scan(&q)
    query := scanStrings(q)

    // fmt.Println(q)
    // fmt.Println(query)

    i := 0
    for i < q {
        //fmt.Println(query[i][0:1])
        if query[i][0:1] == "1" {
            s=Reverse(s)
        } else {
            if query[i][1:2] == "1" {
                s = query[i][2:3] + s
            } else {
                s = s + query[i][2:3]
            }
        }
        i++
    }
    fmt.Println(s)
}

func scanStrings(len int) (strings []string) {
    var str string 
    var f string
    var c string
    for i := 0; i < len; i++ {
        f = ""
        c = ""
        fmt.Scanf("%s", &str)
        if str == "2" {
            fmt.Scanf("%s", &f)
            fmt.Scanf("%s", &c)
        }
        strings = append(strings, str+f+c)
    }
    return
}

func Reverse( s string ) string {
    b := make([]byte, len(s));
    var j int = len(s) - 1;
    for i := 0; i <= j; i++ {
        b[j-i] = s[i]
    }

    return string ( b );
}