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: Time Limit Exceeded)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;

public class Main {
    public static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));

    public static int nextInt() {
        try {
            in.nextToken();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        return (int) in.nval;
    }

    public static void main(String[] args) throws IOException {
        // TODO Auto-generated method stub
        in.nextToken();
        StringBuffer s = new StringBuffer(in.sval);
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int n = nextInt(), a = 0;
        for (int i = 0; i < n; i++) {
            int op = nextInt();
            if (op == 1)
                a++;
            else {
                int x = nextInt();
                in.nextToken();
                String c = in.sval;
                int f = a & 1;
                if ((x == 1 && f == 0) || (x == 2) && f == 1)
                    s.insert(0, c);
                else
                    s.append(c);
            }
        }
        int f = a & 1;
        if (f == 0)
            out.println(s);
        else {
            String ss = s.reverse().toString();
            out.print(ss);
        }
        out.flush();
    }

}