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.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    String S = scan.next();
    int Q = scan.nextInt();
    long[] T = new long[Q];
    long[] F = new long[Q];
    String[] C = new String[Q];
    for(int i=0; i<Q; i++) {
      T[i] = scan.nextLong();
      if(T[i] == 2) {
        F[i] = scan.nextLong();
        C[i] = scan.next();
      }
    }
    scan.close();

    String before = "";
    String after = "";
    int count = 0;
    for(int i=0; i<Q; i++) {
      if(T[i] == 1) {
        count++;
      }
      else {
        if((count + F[i]) % 2 == 1) {
          before += C[i];
        }
        else {
          after += C[i];
        }
      }
    }

    if(count % 2 == 1) {
      StringBuffer afterReverse = new StringBuffer(after);
      StringBuffer SReverse = new StringBuffer(S);
      System.out.print(afterReverse.reverse());
      System.out.print(SReverse.reverse());
      System.out.println(before);
    }
    else {
      StringBuffer beforeReverse = new StringBuffer(before);
      System.out.print(beforeReverse.reverse());
      System.out.print(S);
      System.out.println(after);
    }
  }
}