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);
	String S = sc.next();
StringBuilder sbA = new StringBuilder();
StringBuilder sbB = new StringBuilder();
	int Q = sc.nextInt();
	int[][] query = new int[Q][2]; //各命令の１、２つめを格納
	char[] queryChar = new char[Q]; //各命令のcharを格納
	//List<Character> listA = new ArrayList<>();
    //List<Character> listB = new ArrayList<>();
	int revCount =0;
	//２つのリストを作る
	for(int i=0; i<Q; i++) {
		query[i][0] = sc.nextInt();
		if ( query[i][0] == 2) {
			query[i][1] = sc.nextInt();
			queryChar[i] = sc.next().charAt(0);
		}
	}
	
	//命令実行部
	for(int i=0; i<Q; i++) {
		//反転
		if( query[i][0] == 1){
			revCount++;
			//反転していない状態
	} else if( revCount % 2 == 0) {
		    //先頭に追加
			if ( query[i][1] == 1) {
				//listA.add(0,queryChar[i]);
				sbA.insert(0,queryChar[i]);	
				//末尾に追加
			} else if  (query[i][1] == 2){
				//listA.add(queryChar[i]);
				sbB.append(queryChar[i]);
			}
			//反転してる状態
		} else {
			if ( query[i][1] == 1) {
				//listA.add(queryChar[i]);
				sbB.append(queryChar[i]);
			} else if  (query[i][1] == 2){
				//listA.add(0,queryChar[i]);
				sbA.insert(0,queryChar[i]);	
			}			
		}		
	}
	
		if ( revCount %2 == 0)
		//for(int i=0; i< listA.size(); i++)System.out.print(listA.get(i));
			System.out.print(sbA + S + sbB);
		else {
			char[] C = S.toCharArray();
			StringBuilder sbC = new StringBuilder();
			for(int i=0; i<C.length; i++) {
				sbC.append(C[C.length - i - 1]);
			}
			System.out.print(sbA);
			System.out.print(sbC);			System.out.print(sbB);				
		}
			
		//for(int i=0; i<listA.size(); i++) 
			//System.out.print(listA.get(listA.size() - 1 - i));
			
  }
}