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: Time Limit Exceeded)

using System;
using System.Text;
using System.Linq;
using System.Collections;
using System.Collections.Generic;
using static System.Console;
using static System.Math;
using System.Numerics;

namespace AtCoder
{
    public class Program
    {
        public static void Main(string[] args)
        {
            new Program().Solve(new ConsoleInput(Console.In, ' '));
            Console.ReadLine();
        }

        public void Solve(ConsoleInput cin)
        {
            string s = cin.Read;
            int q = cin.ReadInt;
            List<char> p = new List<char>();
            for (int i = 0; i < s.Length; i++)
            {
                p.Add(s[i]);
            }

            bool rever = false;
            int min = 0;
            int max = s.Length - 1;
            for (int i = 0; i < q; i++)
            {
                string c = Console.ReadLine();
                if (int.Parse(c[0].ToString()) == 1)
                {
                    //逆転
                    rever = !rever;
                }
                else
                {
                    string[] sp = c.Split(' ');
                    int f = int.Parse(sp[1]);
                    char mozi = (char)sp[2][0];
                    if ((f == 1 && !rever) || (f == 2 && rever))
                    {
                        //先頭に追加するとき
                        p.Insert(0, mozi);
                    }
                    else
                    {
                        //後ろに追加するとき
                        p.Add(mozi);
                    }
                }
            }

            if (!rever)
            {
                Console.WriteLine(p.ToArray());
            }
            else
            {
                p.Reverse();
                Console.WriteLine(p.ToArray());
            }
        }

        // 以下は消さないこと

        //a^n mod MODを求める(繰り返し2情報を用いた高速版)
        static long modpow(long a, long n, long MOD)
        {
            if (n == 0)
                return 1;
            if (n % 2 == 0)
            {
                long temp = modpow(a, n / 2, MOD);
                return temp * temp % MOD;
            }

            return a * modpow(a, n - 1, MOD) % MOD;
        }

        // 最小公倍数
        public static long Lcm(long a, long b)
        {
            return a * b / Gcd(a, b);
        }

        // ユークリッドの互除法 
        public static long Gcd(long a, long b)
        {
            if (a < b)
                // 引数を入替えて自分を呼び出す
                return Gcd(b, a);
            while (b != 0)
            {
                var remainder = a % b;
                a = b;
                b = remainder;
            }
            return a;
        }
    }

    public class Vector2
    {
        public int x;
        public int y;

        public Vector2(int _x, int _y)
        {
            x = _x;
            y = _y;
        }
    }
    public class ConsoleInput
    {
        private readonly System.IO.TextReader _stream;
        private char _separator = ' ';
        private Queue<string> inputStream;
        public ConsoleInput(System.IO.TextReader stream, char separator = ' ')
        {
            this._separator = separator;
            this._stream = stream;
            inputStream = new Queue<string>();
        }
        public string Read
        {
            get
            {
                if (inputStream.Count != 0) return inputStream.Dequeue();
                string[] tmp = _stream.ReadLine().Split(_separator);
                for (int i = 0; i < tmp.Length; ++i)
                    inputStream.Enqueue(tmp[i]);
                return inputStream.Dequeue();
            }
        }
        public string ReadLine { get { return _stream.ReadLine(); } }
        public int ReadInt { get { return int.Parse(Read); } }
        public long ReadLong { get { return long.Parse(Read); } }
        public ulong ReadULong { get { return ulong.Parse(Read); } }
        public double ReadDouble { get { return double.Parse(Read); } }
        public string[] ReadStrArray(long N) { var ret = new string[N]; for (long i = 0; i < N; ++i) ret[i] = Read; return ret; }
        public int[] ReadIntArray(long N) { var ret = new int[N]; for (long i = 0; i < N; ++i) ret[i] = ReadInt; return ret; }
        public long[] ReadLongArray(long N) { var ret = new long[N]; for (long i = 0; i < N; ++i) ret[i] = ReadLong; return ret; }
        public ulong[] ReadULongArray(long N) { var ret = new ulong[N]; for (long i = 0; i < N; ++i) ret[i] = ReadULong; return ret; }
    }

    public class Util
    {
        //nCrを計算したいときに使用
        public static ModInt Comb(int n, int r)
        {
            ModInt res = 1;
            for (int i = 1; i <= r; i++)
            {
                res *= n;
                res /= i;
                n--;
            }
            return res;
        }

        //n^mを計算
        public static ModInt Power(ModInt n, long m)
        {
            ModInt pow = n;
            ModInt res = 1;
            while (m > 0)
            {
                if ((m & 1) == 1) res *= pow;
                pow *= pow;
                m >>= 1;
            }
            return res;
        }
    }


    public struct ModInt
    {
        public const int Mod = 1000000007;
        const long POSITIVIZER = ((long)Mod) << 31;
        long Data;
        public ModInt(long data) { if ((Data = data % Mod) < 0) Data += Mod; }
        public static implicit operator long(ModInt modInt) => modInt.Data;
        public static implicit operator ModInt(long val) => new ModInt(val);
        public static ModInt operator -(ModInt a, int b) => new ModInt() { Data = (a.Data - b + POSITIVIZER) % Mod };
        public static ModInt operator -(ModInt a, ModInt b) { long res = a.Data - b.Data; return new ModInt() { Data = res < 0 ? res + Mod : res }; }
        public static ModInt operator *(ModInt a, int b) => new ModInt(a.Data * b);
        public static ModInt operator *(ModInt a, ModInt b) => new ModInt() { Data = a.Data * b.Data % Mod };
        public static ModInt operator /(ModInt a, ModInt b) => new ModInt() { Data = a.Data * GetInverse(b) % Mod };
        public override int GetHashCode() => (int)Data;
        static long GetInverse(long a)
        {
            long div, p = Mod, x1 = 1, y1 = 0, x2 = 0, y2 = 1;
            while (true)
            {
                if (p == 1) return x2 + Mod; div = a / p; x1 -= x2 * div; y1 -= y2 * div; a %= p;
                if (a == 1) return x1 + Mod; div = p / a; x2 -= x1 * div; y2 -= y1 * div; p %= a;
            }
        }
    }
}