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)

#include "bits/stdc++.h"

using namespace std;
typedef long long int ll;
typedef pair<ll, ll > pi;  
typedef pair<pair<ll, ll >, ll > pii;  
vector<ll > vec;
vector<vector<ll > > vec2;
ll MOD = 1000000007;
ll INF = 1145141919454519;

int main(){

    string S;
    cin >> S;

    ll Q;
    cin >> Q;

    ll rev = 0;
    deque<char > deq;
    for(ll i = 0; i < S.length(); i++){
        deq.push_front(S[i]);
    }

    for(ll i = 0; i < Q; i++){
        ll t; cin >> t;
        if(t == 1){
            rev++;
        }else{
            ll f; cin >> f;
            char c; cin >> c;
            if(f == 1){
                if(rev % 2 == 0){
                    deq.push_front(c);
                }else{
                    deq.push_back(c);
                }
            }else{
                if(rev % 2 == 0){
                    deq.push_back(c);
                }else{
                    deq.push_front(c);
                }
            }
        }
    }

    ll N = deq.size();
    for(ll i = 0; i < N; i++){
        if(rev % 2 == 0){
            cout << deq.front();
            deq.pop_front();
        }else{
            cout << deq.back();
            deq.pop_back();
        }
    }
    cout << endl;
    

}