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.lang.*; 
import java.io.*; 
import java.util.*; 
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;


public class Main {
	static class FastReader{
		BufferedReader br;
		StringTokenizer st;
		public FastReader() {
			br =  new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {
			while(st==null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
					
				}
				catch(IOException e) {
					e.printStackTrace();
				}
				
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	static String add_string(String s1,String s2) {
		return s1+s2;
	}
	public static void main(String[] args) {
		FastReader ob = new FastReader();
		String s = ob.next();
		boolean r =false;
		int q = ob.nextInt();
		for(int i=0;i<q;i++) {
			int t = ob.nextInt();
			if(t==1) {
				r=true;
			}
			else {
				int f =ob.nextInt();
				String s1=ob.next();
				if(r) {
					if(f==2) {
						s=add_string(s1, s);
					}
					else {
						s=add_string(s, s1);
					}
					r=false;
				}
				else {
					if(f==2) {
						s=add_string(s, s1);
					}
					else {
						s=add_string(s1, s);
					}
					
				}
			}
		}
		char arr[] = s.toCharArray();
		if(r) {
			for (int i = arr.length-1; i >-1; i--) {
				System.out.print(arr[i]);
			}
		}
		else {
			System.out.println(s);
		}
	}

}