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.Objects;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        String S = s.next();
        int Q = Integer.parseInt(s.next());
        StringBuilder origin = new StringBuilder(S);
        StringBuilder front = new StringBuilder();
        StringBuilder back = new StringBuilder();
        /**
         * 4 2 1 p 1 2 2 c 1
         */
        /**
         * 123 先頭にp追加 p123 反転 321p 末尾にc追加 321pc 反転 cp123 先頭にa追加 acp123 末尾にb追加 acp123b
         */
        // 反転しているかどうか
        // 反転していない場合の先頭追加、末尾追加
        // 反転している場合の先頭追加、末尾追加
        boolean normal = true;
        for (int i = 0; i < Q; i++) {
            String type = s.next();
            // 反転
            if (Objects.equals(type, "1")) {
                normal = !normal;
                continue;
            }
            // 追加処理
            String F = s.next();
            String C = s.next();
            if (Objects.equals(F, "1")) {
                // 先頭追加処理
                if (normal) {
                    front.append(C);
                } else {
                    back.append(C);
                }
            } else {
                // 末尾追加処理
                if (normal) {
                    back.append(C);
                } else {
                    front.append(C);
                }
            }
        }
        // 表示処理
        if (normal) {
            System.out.print(front.reverse());
            System.out.print(origin);
            System.out.println(back);
        }else{
            System.out.print(back.reverse());
            System.out.print(origin.reverse());
            System.out.println(front);
        }
    }
}