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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)


import java.util.Scanner;

public class Main {

     static Scanner scanner;

     public static void main(String[] args) {
         scanner = new Scanner(System.in);

         String S=gs();
         int Q=gi();
         int[] T=new int[Q];
         int[] F=new int[Q];
         String[] C=new String[Q];

         for (int i=0; i<Q;i++) {
        	 T[i]=gi();
        	 if(T[i]==2) {
        		 F[i]=gi();
        		 C[i]=gs();
        	 }
         }
         
         StringBuilder sb= new StringBuilder(S);
         boolean rev=false;
         for (int i=0; i<Q;i++) {
        	 if(T[i]==1) {
        		 rev=!rev;
        	 } else {
        		 if((rev==false && F[i]==1) || rev==true && F[i]==2) {
        			 sb.insert(0,C[i]);
        		 } else {
        			 sb.append(C[i]);
        		 }
        	 }
         }
         StringBuffer sbf=new StringBuffer(sb);
         if(rev)sbf.reverse();

         System.out.println(sbf.toString());
     }

     // 文字列として入力を取得
     public static String gs() {
          return scanner.next();
     }

     // intとして入力を取得
     public static int gi() {
          return Integer.parseInt(scanner.next());
     }

     // longとして入力を取得
     public static long gl() {
          return Long.parseLong(scanner.next());
     }

     // doubleとして入力を取得
     public static double gd() {
          return Double.parseDouble(scanner.next());
     }
}