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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections;
using System.Collections.Generic;
using System.Text;

class Program
{
    static void Main(string[] args)
    {
        // 入力
        string s = Console.ReadLine();
        long q = long.Parse(Console.ReadLine());
        string[][] jagQ = new string[q][];
        for (long i = 0; i < jagQ.Length; i++)
        {
            jagQ[i] = Console.ReadLine().Split().ToArray();
        }

        var list = new LinkedList<string>();
        for (int i = 0; i < s.Length; i++)
        {
            list.AddLast(s[i].ToString());
        }

        var rev = false;
        for (long i = 0; i < q; i++)
        {
            if (jagQ[i][0] == "1")
            {
                rev = !rev;
            }
            else
            {
                var tmp = int.Parse(jagQ[i][1]);
                // 反転状態だと 1->2, 2->1 にして追加先を変更する
                if (rev) tmp = (tmp + 1) % 2;

                if (tmp == 1)
                {
                    list.AddFirst(jagQ[i][2]);
                }
                else
                {
                    list.AddLast(jagQ[i][2]);
                }
            }
        }

        // 反転状態の場合は後ろから出力するのを忘れない
        if (rev)
        {
            foreach (var item in list.Reverse())
            {
                Console.Write(item);
            }
        }
        else
        {
            foreach (var item in list)
            {
                Console.Write(item);
            }
        }
        Console.WriteLine();

    }
}