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>
using namespace std;
 
 
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using ll = long long;
 
/* vector<int> A(N); 
    vector<pair<int, int> > tasks; ＃ソートした場合、１つ目でソート
　tie(b, a) = p.at(i); #pairを分割
    tasks.emplace_back(b, a)  ＃配列tasksの最後に(b,a)を追加
　xの二乗:  pow(x, 2.0) 
    小さい順: sort(v.begin(), v.end());
    大きい順: reverse(v.begin(), v.end());
    int: -2,147,483,648 ～ 2,147,483,647
    ll: -9,223,372,036,854,775,808 ~ 9,223,372,036,854,775,807
    // int型の2次元配列(3×4要素の)の宣言
    vector<vector<int>> data(3, vector<int>(4));
    reverse(str.begin(), str.end());

template <class BidirectionalIterator>
void reverse(BidirectionalIterator first, BidirectionalIterator last) {
  for ( ; first != last && first != --last; ++first)
    std::iter_swap(first, last);
}
*/

int main() {
  
  ll A, B, C,Q , count=0, ans=-1;
  string S;
  cin >> S >> Q;
  
  vector<int> in(Q); 
  vector<int> inin(Q); 
  vector<char> ch(Q); 

  rep(i,Q){
  	cin >> in.at(i);
    if(in.at(i)==1){
      reverse(S.begin(), S.end());
    }
    
    else if(in.at(i)==2){
      cin >> inin.at(i) >> ch.at(i);
      if(inin.at(i)==1){
       S=ch.at(i)+S;
      }
      else if(inin.at(i)==2){
       S=S+ch.at(i);
      }
      
    }
  }
  
  cout << S << endl;
}