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.Linq;
using System.Collections.Generic;
 
class Program
{
	static void Main(string[] args)
	{
		//読み込む
		string S = Console.ReadLine();
		int Q = int.Parse(Console.ReadLine());
 
		//反転回数が奇数かを確かめる変数
		bool isOdd = true;
 
		//ループ処理で読み込む
		for(int i=0; i<Q; i++){
				string[] line = Console.ReadLine().Split(' ');
				string Ti = line[0];
				//Ti = 1のとき
				if(Ti == "1"){
						//文字列の長さが二以上の場合のみ、反転回数を加算
						if(S.Length > 1){
								if(isOdd){
										isOdd = false;
								}else{
										isOdd = true;
								}
						}
				}
 
				//Ti = 2のとき
				else{
						//命令を変数に格納
						int Fi = int.Parse(line[1]);
						string Ci = line[2];
 
						//文字列の長さを取得
						int len = S.Length;
 
						//反転回数が偶数の場合
						if(isOdd){
								if(Fi == 1){
										//先頭に追加
										S = Ci + S;
								}else{
										//末尾に追加
										S += Ci;
								}
						}
						//奇数の場合
						else{
								if(Fi == 2){
										//先頭に追加
										S = Ci + S;
								}else{
										//末尾に追加
										S += Ci;;
								}
						}
				}
		}
 
		//反転回数が偶数の場合、そのまま出力
		if(isOdd){
				Console.WriteLine(S);
		}
		/*/反転回数が奇数の場合、文字列を逆から出力
		else{
        int len = S.Length;
        for(int i=0; i<len; i++){
            Console.Write(S.Substring(len - i - 1, 1));
        }
		}
		/*/
	}
}