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: Wrong Answer)

///mAAria...
#include<bits/stdc++.h>
 
using namespace std;
 
#define int long long
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define all(v) v.begin(),v.end()
#define ld long double
#define ff first
#define ss second
#define pii pair<int,int>
#define pi 3.1415926535897932384626433832
#define mod 1000000007
#define inf 1000000000
#define pr1(a) cout << a << endl
#define pr2(a,b)cout << a <<  " " <<b<< endl
#define pr3(a,b,c) cout << a <<" " << b << " " << c << endl

const long double PI = acos(-1);

int powm(int a, int b) {
    int res=1;
    while(b) {
        if(b&1)
            res=(res*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return res;
}

int gcd(int a, int b){
    if(b==0)return a;
    gcd(b,a%b);
}


void solve(){
    
    string s;
    cin >> s;
    int m = s.size(), l =0;

    int n;
    cin >> n;
    int rev = 0;
    while(n--){
        int t;
        cin >> t;

        if(t==2){
            int f;
            char c;
            cin >> f >> c;
            f--;

            if(rev%2){
                if(f){
                    s = c + s;
                    l++;
                }
                else {
                    s = s + c;
                    
                }
            }
            else{
                if(!f){
                    s = c + s;
                    l++;
                }
                else {
                    s = s + c;
                    
                }
               
            }
        }
        else 
            rev++;
    }

    if(rev%2){
        reverse(s.begin()+l, s.begin()+l+m);
        cout << s << endl;
    }
    else cout << s << endl;

 
} 

signed main() 
{   
    IOS;
    int t = 1;
    //cin >> t;

    for(int i = 1;i<=t;i++){
        solve();
    }
    return 0;
}