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: Time Limit Exceeded)

#include <bits/stdc++.h>          
using namespace std;               
typedef long long ll;

int main() {
    string S;
    ll Q;
    cin >> S >> Q;

    vector<int> T(Q), F(Q);
    vector<char> C(Q);
    for (ll i=0; i<Q; i++) {
        cin >> T[i];
        if (T[i] == 1) {
            F[i] = 0;
            C[i] = ' ';
        }
        else {
            cin >> F[i] >> C[i];
        }
    }

    //
    ll T1count=0, F1count=0, F2count=0;
    for (ll i=0; i<Q; i++) {
        if (T[i] == 1) {
            T1count++;
        }
        else if (F[i] == 1) {
            if (T1count%2 == 0){
                S = C[i] + S;
            }
            else {
                S = S + C[i];
            }
        }
        else {
            if (T1count%2 == 0) {
                S = S + C[i];
            }
            else {
                S = C[i] + S;
            }
        }
    }
    
    if (T1count%2 != 0) {
        reverse(S.begin(),S.end());
    }
    
    cout << S << endl;
}