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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;

int debug = 0;

int main()
{
    string s;
    int q;
    int n, f;
    string c; 

    cin >> s;
    cin >> q;
    vector<tuple<int, int, string>> op;

    rep(i,q){
        cin >> n;
        if(n==1){
            op.emplace_back(1, -1, "");
        }else{
            cin >> f >> c;
            op.emplace_back(2, f, c);
        }
    }    

    if(debug){
        for(auto x:op){
            cout << "<" << get<0>(x) << "," << get<1>(x) << "," << get<2>(x) << ">";
            cout << endl;
        }
    }

    if(debug){
        cout << s << endl;
    }
    for(auto x:op){
        if(get<0>(x)==1){ // reverse
            reverse(s.begin(), s.end());
            if(debug)
                cout << "r:" << s << endl;
        }else if(get<1>(x)==1){ // head
            s = get<2>(x)+s;
            if (debug)
                cout << "h:" << get<2>(x) << ":" << s << endl;
        }
        else if (get<1>(x) == 2)
        { // tail
            s = s+get<2>(x) ;
            if (debug)
                cout << "t:" << get<2>(x) <<":"<< s << endl;
        }
    }

    cout << s << endl;

    return 0;
}