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.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main{
	static BufferedReader br;
	static int cin() throws Exception
	{
		return Integer.valueOf(br.readLine());
	}
	static int[] split() throws Exception
	{
		String[] cmd=br.readLine().split(" ");
		int[] ans=new int[cmd.length];
		for(int i=0;i<cmd.length;i++)
		{
			ans[i]=Integer.valueOf(cmd[i]);
		}
		return ans;
	}
	static long[] splitL() throws IOException
	{
		String[] cmd=br.readLine().split(" ");
		long[] ans=new long[cmd.length];
		for(int i=0;i<cmd.length;i++)
		{
			ans[i]=Long.valueOf(cmd[i]);
		}
		return ans;
	}
	public static void main(String[] args) throws Exception{
		// TODO Auto-generated method stub
		br=new BufferedReader(new InputStreamReader(System.in));
		String s=br.readLine();
		int q=cin();
		int rev=0;
		for(int i=0;i<q;i++)
		{
			String[] st=br.readLine().split(" ");
			int t=Integer.valueOf(st[0]);
			if(t==1)
				rev=rev^1;
			else
			{
				int f=Integer.valueOf(st[1]);
				if(f==2)
				{
					if(rev==0)
						s=s+""+st[2];
					else
						s=st[2]+s;
				}
				else
				{
					if(rev==0)
						s=st[2]+""+s;
					else
						s=s+""+st[2];
				}
			}
		}
		if(rev==0)
			System.out.println(s);
		else
		{
			StringBuilder sr=new StringBuilder(s);
			System.out.println(sr.reverse().toString());
		}
	}
}