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: Wrong Answer)

#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<utility>
#include<map>
#include<set>
#include <sstream>
#include<queue>
#include<stack>
#include<functional>
#include<math.h>
#include <iomanip>
#include <numeric>
#include <iterator>
#include <math.h>
#include <list>
using namespace std;

int main(void){
  string s;

  cin >> s;
  int Q;
  cin >> Q;
  int dir = 1;
  list<char>neko;
  for(int i =0; i < s.size(); i++){
      neko.push_back(s[i]);
  }
  for(int i =0; i < Q; i++){
    int T1;
    cin >> T1;
    switch(T1)
    {
    case 1:
      dir *= -1;
      break;
    case 2:
      int T2;
      cin >> T2;
      char c;
      cin >> c;
      if(T2 == 2){
        if(dir == -1){
          neko.push_front(c);
        }else{
          neko.push_back(c);
        }
      }else{
        if(dir == -1){
          neko.push_back(c);
        }else{
          neko.push_front(c);
        }
      }
      break;
    }
  }
  cout << dir << endl;
  if(dir == -1){
    reverse(neko.begin(), neko.end());
  }
  for(auto it = neko.begin(); it != neko.end(); ++it){
    cout << *it;
  }
  cout << endl;
  return 0;
}