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 <algorithm>
#include <cmath>
#include <ctime>
#include <iostream>
#include <cstdlib>
#include <string>
#define sbc __builtin_popcount
#define pb push_back
#define em emplace
#define emb emplace_back
int mod = 1e9+7;
using namespace std;
    void solve() {
                string str,b="",e="";
                cin>>str;
                char ch;
                int q,r=0,t,f;
                cin>>q;
                while(q--) {
                        cin>>t;
                        if(t-1) {
                                cin>>f>>ch;
                                if(f-1) {
                                        if(r) {
                                                b=ch+b;
                                        } else {
                                                e+=ch;
                                        }
                                } else {
                                        if(r) {
                                                e+=ch;
                                        } else {
                                                b=ch+b;
                                        }
                                }
                        } else {
                                r=1-r;
                        }
                }
                str=b+str+e;
                if(r) {reverse(str.begin(),str.end());}
                cout<<str<<'\n';
    }

int main()
{

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);



   // int t;
   // cin>>t;
   // while(t--)
          solve();


    return 0;
}