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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <map>
#include <set>
#include <stack>
#include <tuple>
#include <deque>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cassert>
#include <chrono>
#include <random>
#include <limits>
#include <iterator>
#include <functional>
#include <sstream>
#include <complex>
using namespace std;

#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)

typedef long long ll;
typedef pair<int, int> P;
typedef pair<int, double> Pid;
typedef pair<double, int> Pdi;
typedef pair<ll, int> Pl;
typedef pair<int, pair<int, int>> PP;
constexpr double PI = 3.1415926535897932;   // acos(-1)
constexpr double EPS = 1e-9;
constexpr int INF = 1001001001;
constexpr int mod = 1e+9 + 7;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    string s;
    cin >> s;
    int n = s.length();
    deque<char> deq;
    for(int i = 0; i < n; ++i)  deq.push_back(s[i]);
    int q;
    cin >> q;
    int flag = 0;
    for(int i = 0; i < q; ++i){
        int t;
        cin >> t;
        if(t == 1){
            flag ^= 1;
        }
        else{
            int f;
            char c;
            cin >> f >> c;
            if(flag == 0){
                if(f == 1){
                    deq.push_front(c);
                }
                else{
                    deq.push_back(c);
                }
            }
            else{
                if(f == 1){
                    deq.push_back(c);
                }
                else{
                    deq.push_front(c);
                }
            }
        }
    }
    if(flag == 0){
        while(!deq.empty()){
            cout << deq.front();
            deq.pop_front();
        }
        cout << endl;
    }
    else{
        while(!deq.empty()){
            cout << deq.back();
            deq.pop_back();
        }
        cout << endl;
    }
}