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 <iostream>
#include <cfenv>
#include <cmath>
#include <vector>
#include <algorithm>
#include <string>
#include <deque>

using namespace std;
using ll = long long;

int main(){
  bool reverse_flag = false;
  char s;
  cin >> s;

  deque<char> S;
  S.emplace_back(s);

  int Q;
  cin >> Q;

  for(int i = 0; i < Q; ++i){
    int T;
    cin >> T;
    if(T == 1){
      reverse_flag = !reverse_flag;
    } else if(T == 2){
      int F;
      char C;

      cin >> F >> C;

      if(reverse_flag){
        if(F == 1){
          S.emplace_back(C);
        }else if(F == 2){
          S.emplace_front(C);
        }
      } else{
        if(F == 1){
          S.emplace_front(C);
        }else if(F == 2){
          S.emplace_back(C);
        }
      }
    }
  }

  string ans = "";

  for_each(begin(S), end(S), [&ans](char v) {
    ans += v;
  });

  if(reverse_flag){
    reverse(ans.begin(), ans.end());
  }
  
  cout << ans << endl;

  return 0;
}