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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

typedef struct TFC
{
    int T;
    int F;
    string C;
};

void solve(std::istream &ist, std::ostream &ost)
{
    /* ------- 宣言と入力 ------- */

    string S;
    int Q;

    ist >> S;
    ist >> Q;

    struct TFC Q_uery[Q];

    for (int i = 0; i < Q; i++)
    {
        ist >> Q_uery[i].T;
        if (Q_uery[i].T == 2)
        {
            ist >> Q_uery[i].F;
            ist >> Q_uery[i].C;
        }
    }

    /* ------- 計算 ------- */

    bool mode = true;
    deque<string> deq;

    for (int i = 0; i < S.length(); i++)
    {
        deq.emplace_back(S.substr(i, 1));
    }

    for (int i = 0; i < Q; i++)
    {
        if (Q_uery[i].T == 1)
        {
            mode = !mode;
        }
        else
        {
            if ((Q_uery[i].F == 1 && mode) || (Q_uery[i].F == 2 && !mode))
            {
                deq.emplace_front(Q_uery[i].C);
            }
            else if ((Q_uery[i].F == 1 && !mode) || (Q_uery[i].F == 2 && mode))
            {
                deq.emplace_back(Q_uery[i].C);
            }
        }
    }

    if (mode)
    {
        for (auto it = deq.begin(); it != deq.end(); ++it)
        {
            ost << *it;
        }
        ost << endl;
    }
    else
    {
        for (auto it = deq.end() - 1; it == deq.begin(); --it)
        {
            ost << *it;
        }
        ost << endl;
    }
}

#ifndef WOMAIN
int main()
{
    solve(std::cin, std::cout);
    return 0;
}
#endif