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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*

fun main(args:Array<String>){
    solve()
}

fun solve(){
    val s = read().toCharArray()
    val ans = ArrayDeque<Char>()
    s.forEach{
        ans.addLast(it)
    }
    var dir = true
    val q = readint()
    repeat(q){
        val a = reads()
        if(a[0] == "1"){
            dir = !dir
        }
        else{
            val f = a[1].toInt()
            if(f == 1){
                if(dir){
                    ans.addFirst(a[2][0])
                }
                else{
                    ans.addLast(a[2][0])
                }
            }
            else{
                if(!dir){
                    ans.addFirst(a[2][0])
                }
                else{
                    ans.addLast(a[2][0])
                }
            }
        }
    }
    val res = StringBuilder()
    if(dir){
        while(ans.isNotEmpty()){
            res.append(ans.first)
            ans.removeFirst()
        }
    }
    else{
        while(ans.isNotEmpty()){
            res.append(ans.last)
            ans.removeLast()
        }
    }
    println(res)
}

fun read() = readLine()!!
fun reads() = readLine()!!.split(" ")
fun readint() = readLine()!!.toInt()
fun readints() = readLine()!!.split(" ").map{it.toInt()}
fun readlong() = readLine()!!.toLong()
fun readlongs() = readLine()!!.split(" ").map{it.toLong()}