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: Wrong Answer)

import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		// 整数の入力
      String S = sc.next();
      int Q = sc.nextInt();
      ArrayList<String> bf = new ArrayList<>();
      ArrayList<String> af = new ArrayList<>();
      boolean asc =true;
      int Ti;
      int Fi;
      String Ci;
	for(int i=0;i<Q;i++){
      Ti = sc.nextInt();
      if(Ti==1){
        asc = !asc;
        continue;
      }
      Fi = sc.nextInt();
      if(asc){
        if(Fi==1){
          bf.add(sc.next());
        }else{
          af.add(sc.next());
        }
      }else{
        if(Fi==2){
          bf.add(sc.next());
        }else{
          af.add(sc.next());
        }
      }
    }
      String str ="";
if(asc){
  Collections.reverse(bf);
  for(String b:bf){
    str+=b;
  }
  str+=S;
  for(String a:af){
    str+=a;
  }
}else{
    Collections.reverse(af);
  for(String a:af){
    str+=a;
  }
  str+=S;
  for(String b:bf){
    str+=b;
  }
}
      System.out.println(str);


    }
  
  private static long lcm(long m, long n) {
    return m * n / gcd(m, n);
}
  
  private static long gcd(long m, long n) {
    if(m < n) return gcd(n, m);
    if(n == 0) return m;
    return gcd(n, m % n);
}
}