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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;
import java.io.*;

public class Main{
  public static void main(String[] args)throws Exception{
    Scanner sc = new Scanner(System.in);
    String S = sc.nextLine();
    Deque<String> deq = new ArrayDeque<>();
    deq.add(S);
    int N = Integer.parseInt(sc.nextLine());
    boolean f = true;
    for(int i = 0; i < N; i++){
      String[] T = sc.nextLine().split(" ");
      int order = Integer.parseInt(T[0]);
      if(order == 1){
        f = !f;
      }else{
        int o2 = Integer.parseInt(T[1]);
        if(o2 == 1){
          if(f){
            deq.addFirst(T[2]);
          }else{
            deq.addLast(T[2]);
          }
        }else{
          if(f){
            deq.addLast(T[2]);
          }else{
            deq.addFirst(T[2]);
          }
        }
      }
    }
    StringBuilder sb = new StringBuilder();
    for(String s : deq){
      sb.append(s);
    }
    System.out.println(f ? sb.toString() : sb.reverse().toString());
  }
}