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.*;
public class Main {
	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		
		String s = sc.nextLine();
		char[]arr = new char[300001];
		for(int i = 0 ; i<s.length(); i++) {
			arr[i] = s.charAt(i);
		}
		int arrindex = s.length();
		char[]front = new char[200001];
		int frontindex = 0;
		int Q = sc.nextInt();
		boolean reverse = false;
		while(Q-->0) {
			int op = sc.nextInt();
			if(op == 1) {
				reverse = !reverse;
			}
			else {
				int pos = sc.nextInt();
				char add = sc.next().charAt(0);
				if(pos == 1 && !reverse) {
					front[frontindex] = add;
					frontindex++;
				}
				else if(pos == 1 && reverse) {
					arr[arrindex] = add;
					arrindex++;
			//		System.out.println("add " + add);
				}
				else if(pos == 2 && !reverse) {
				//	System.out.println("add " + add);
					arr[arrindex] = add;
					arrindex++;
				}
				else if(pos == 2 && reverse) {
					front[frontindex] = add;
					frontindex++;
				}
			}
		}
	//	String res = "";
		if(!reverse) {
			for(int i = frontindex-1; i>=0 ; i--) {
				System.out.print(front[i]);
		//		res+=front[i];
			}
			for(int i = 0; i<arrindex; i++) {
				System.out.print(arr[i]);
				//res+=arr[i];
			}
		}
		else {
			for(int i = arrindex-1 ; i>=0; i--) {
				//res+=arr[i];
				System.out.print(arr[i]);
			}
			for(int i = 0; i<frontindex; i++) {
				//res+=front[i];
				System.out.print(front[i]);
			}
		}
		System.out.println();
	}
}