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 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define endl "\n"
#define F first
#define S second
#define def(x) cerr << #x << " is " << x << endl;
#define mod 1000000007
#define pb push_back
#define FOR(i,a,n) for(ll i=a;i<n;i++)
#define ios ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define all(a) a.begin(),a.end()
const int NUM = 2e5 + 5;
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }

void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << endl;
    err(++it, args...);
}

int main(){
    ios
    int num_tests=1; 
    //cin>>num_tests;
    while(num_tests-->0){
        string s; cin>>s;
        int q; cin>>q;
        bool cur = false;
        string front,back;
        while(q--){
            int t; cin >> t;
            if(t==1) cur = !cur;
            else{
                int where; cin >> where;
                char what; cin >> what;
                if(where==1 && !cur || where==2 && cur){
                    front.pb(what);
                }
                else back.pb(what);
            }
        }
        if(cur){
            reverse(all(back));
            cout<<back;
            reverse(all(s));
            cout<<s;
            cout<<front;
        }
        else{
            reverse(all(front));
            cout<<front<<s<<back;

        }
    }   
}