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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ULL;
typedef long long LL;
#define FOR(i, a, b) for(LL i=(LL)(a);i<(LL)(b);i++)
#define REP(i,x) for(LL i=0;i<(LL)(x);i++)

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    // cout << fixed << setprecision(10);

    string S;
    cin >> S;
    LL Q;
    cin >> Q;

    bool rev_flag = false;
    REP(i, Q)
    {
        int T;
        cin >> T;
        switch(T)
        {
            case 1:
            {
                // reverse(S.begin(), S.end());
                rev_flag = !rev_flag;
                break;
            }

            case 2:
            {
                LL F;
                char C;
                cin >> F >> C;
                string new_str{C};
                if(rev_flag)
                {
                    F = 3 - F;
                }
                if(F == 1)
                {
                    S = new_str + S;
                }
                else
                {
                    S = S + new_str;
                }
                break;
            }
        }
    }
    if(rev_flag)
    {
        for(auto it=S.end(); it != S.begin(); it--)
        {
            auto ch_temp = it - 1;
            cout << *ch_temp;
        }
        cout << "\n";
    }
    else
    {
        cout << S << endl;
    }
    return 0;
}