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

#include <deque>
#include <string>
#include <iostream>

using namespace std;

int main(int argc, char** argv) {
    string s;
    cin >> s;
    int q;
    cin >> q;

    int status = 0;
    deque<char> s_deque(s.begin(), s.end());
    for (int i = 0; i < q; i++) {
        int t;
        cin >> t;
        if (t == 1) {
            status++;
            continue;
        }
        else {
            int f;
            char c;
            cin >> f;
            cin >> c;
            if (status % 2 == 0) {
                if (f == 1) {
                    s_deque.push_front(c);
                }
                else {
                    s_deque.push_back(c);
                }
            }
            else {
                if (f == 1) {
                    s_deque.push_back(c);
                }
                else {
                    s_deque.push_front(c);
                }
            }
        }
    }

    string res;
    if (status % 2 == 1) {
        res = string(s_deque.rbegin(), s_deque.rend());
    }
    else {
        res = string(s_deque.begin(), s_deque.end());
    }

    cout << res << endl;

    return 0;
}