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.*;

//update 2020/1/1 23:01

public class Main{

  static Scanner sc = new Scanner(System.in);
  static StringBuilder sb = new StringBuilder();

  public static void main(String[] args) {

    String S = next();
    int Q = nextInt();

    sb.append(S);

    for(int i=0;i<Q;i++){
      int T = nextInt();
      if(T==1){
        sb.reverse();
      }
      else{
        int F = nextInt();
        String C = next();
        if(F==1){
          sb.insert(0,C);
        }
        else{
          sb.append(C);
        }
      }
    }


    System.out.println(sb.toString());

  }


  static String toSmall(String s){
    return s.toLowerCase();
  }

  static String toBig(String s){
    return s.toUpperCase();
  }

  static String next(){
    return sc.next();
  }

  static int nextInt(){
    return Integer.parseInt(sc.next());
  }

  static long nextLong(){
    return Long.parseLong(sc.next());
  }

  static double nextDouble(){
    return Double.parseDouble(sc.next());
  }

  static String[] splitString(String s){
    return s.split("");
  }

  static char[] splitChar(String s){
    return s.toCharArray();
  }

  static int charToInt(char a){
    return a - 48;
  }

////////////////////////////////////////////
  public static int max(int[] a){
    int lng = a.length;
    int max = a[0];
    for(int i=1;i<lng;i++){
      max = Math.max(max,a[i]);
    }
    return max;
  }
  public static long max(long[] a){
    int lng = a.length;
    long max = a[0];
    for(int i=1;i<lng;i++){
      max = Math.max(max,a[i]);
    }
    return max;
  }
////////////////////////////////////////////

////////////////////////////////////////////
  public static int min(int[] a){
    int lng = a.length;
    int min = a[0];
    for(int i=1;i<lng;i++){
      min = Math.min(min,a[i]);
    }
    return min;
  }
  public static long min(long[] a){
    int lng = a.length;
    long min = a[0];
    for(int i=1;i<lng;i++){
      min = Math.min(min,a[i]);
    }
    return min;
  }
////////////////////////////////////////////

}