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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
using namespace std;


class Node {
    public:
        char s;
        Node *prev;
        Node *next;
};


int main()
{
    int q;
    int rev;
    int i;
    int t, f;
    char c;

    char s[100000];

    Node *head;
    Node *tail;
    Node *temp;
    head = new Node;
    tail = head;

    cin >> s;
    head->s = s[0];
    head->prev = NULL;
    head->next = NULL;

    i = 1;
    while(s[i] != '\0'){
        temp = new Node;
        temp->s = s[i];
        tail->next = temp;
        temp->prev = tail;
        temp->next = NULL;
        tail = temp;
        i=i+1;
    }

    cin >> q;
    rev = 0;
    for(i=0;i<q;i++){
        cin >> t;

        if(t==1){
            temp = head;
            head = tail;
            tail = temp;
            if(rev == 0){
                rev = 1;
            }else{
                rev = 0;
            }
        }else{
            cin >> f;
            cin >> c;

            if(f == 1){
                temp = new Node;
                temp->s = c;
                if(rev == 0){
                    head->prev = temp;
                    temp->next = head;
                    temp->prev = NULL;
                    head = temp;
                }else{
                    head->next = temp;
                    temp->prev = head;
                    temp->next = NULL;
                    head = temp;
                }
            }else{
                temp = new Node;
                temp->s = c;
                if(rev == 0){
                    tail->next = temp;
                    temp->prev = tail;
                    temp->next = NULL;
                    tail = temp;
                }else{
                    tail->prev = temp;
                    temp->next = tail;
                    temp->prev = NULL;
                    tail = temp;
                }
            }
        }
    }

    temp = head;
    while(1){
        cout << temp->s;
        if(rev == 0){
            temp = temp->next;
        }else{
            temp = temp->prev;
        }
        if(temp == NULL){
            break;
        }
    }

    return 0;

}