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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)


import java.util.*;

// とりあえず入出力例１～３が通るようなロジック
public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		String result = sc.next();
		int n = Integer.parseInt(sc.next());

		// 入力しながらクエリにこたえていこう
		int reverse = 1;
		for (int i = 0; i < n; i++) {

			if (Integer.parseInt(sc.next()) == 1) {
				reverse *= -1;
			} else {
				int kind = Integer.parseInt(sc.next());
				if ((kind == 1 && reverse == 1) || (kind == 2 && reverse == -1)) {
					result = sc.next() + result;
				} else {
					result = result + sc.next();
				}
			}
		}

		// 最後に今が逆行状態なら結果反転
		if (reverse == -1) {
			String str = result;
			result = "";
			for (int i = 0; i < str.length(); i++) {
				result += str.charAt(str.length() - 1 - i);
			}
		}

		System.out.println(result);

	}

}