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: Accepted)

import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;

public class Main {
	public static void main(String[] args){
		try(Scanner sc = new Scanner(System.in)) {
		
			String s = sc.next();
			int q = sc.nextInt();
			
			List<String> sList = new LinkedList<>();
			List<String> revList = new LinkedList<>();
			boolean isRev = false;
			sList.add(s);
			revList.add(new StringBuilder(s).reverse().toString());
			
			for(int i = 0 ; i < q ; i++ ) {
				int op = sc.nextInt();
				
				if ( op == 1 ) {
					isRev = !isRev;
				}
				else {
					int f = sc.nextInt();
					String c = sc.next();
					
					if ( isRev) {
						if ( f == 1 ) {
							//add rev head
							sList.add(c);
							revList.add(0,c);
						}
						else {
							sList.add(0,c);
							revList.add(c);
						}
					}
					else {
						if ( f == 1 ) {
							//add head
							sList.add(0, c);
							revList.add(c);
						}
						else {
							sList.add(c);
							revList.add(0,c);
						}
					}
					
				}
			}
		
			StringBuilder sb = new StringBuilder();					

			if (isRev) {
				for(String ss : revList) {
					sb.append(ss);
				}
			}
			else {
				for(String ss : sList) {
					sb.append(ss);
				}
			}

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