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: Wrong Answer)

using System;
using System.Collections.Generic;

namespace StringFormation
{
    class Program
    {
        static void Main(string[] args)
        {
            bool hantenFlg = false;
            var first = Console.ReadLine().Split(' ');
            List<string> kotae = new List<string>();
            kotae.Add(first[0]);
            var nagasa = int.Parse(Console.ReadLine().Split(' ')[0]);

            for(int i = 0; i < nagasa; i ++)
            {
                var yarukoto = Console.ReadLine().Split(' ');
                if (yarukoto[0] == "1")
                {
                    hantenFlg = Hanten(hantenFlg);
                }
                else
                {
                    kotae = Tuika(kotae, int.Parse(yarukoto[1]), yarukoto[2], hantenFlg);
                }
            }

            string kotaeStr = "";

            if (hantenFlg)
            {
                for(int a = 0; a < kotae.Count; a++)
                {
                    kotaeStr += kotae[kotae.Count - 1 - a];
                }
            }
            else
            {
                foreach (string a in kotae)
                {
                    kotaeStr += a;
                }
            }
            Console.WriteLine(kotaeStr);
        }

        static bool Hanten(bool hantenFlg)
        {
            return !hantenFlg;
        }

        static List<string> Tuika(List<string> taisho, int sentoFlg, string moji, bool hantenFlg)
        {
            if(hantenFlg)
            {
                // 反転の状態
                if (sentoFlg == 1)
                {
                    // 末尾に追加
                    taisho.Add(moji);
                }
                else
                {
                    // 先頭に追加
                    taisho.Insert(0, moji);
                }
            }
            else
            {
                // 通常の状態
                if (sentoFlg == 1)
                {
                    // 先頭に追加
                    taisho.Insert(0, moji);
                }
                else
                {
                    // 末尾に追加
                    taisho.Add(moji);
                }
            }

            return taisho;
        }
    }
}