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;
using ll = long long;

int main(){
  string s;
  int q;
  cin >> s >> q;
  string s1=s.substr(0,s.size()/2);
  string s2=s.substr(s.size()/2);
  int count = 0;
  vector<int> t(q,0);
  vector<int> f(q,0);
  vector<char> c(q);
  for(int i = 0; i < q; i++){
      cin >> t[i];
      if(t[i] == 1){
          count++;
      }else{
          cin >> f[i] >> c[i];
          if(count % 2 == 0){
                if(f[i] == 1){
                    reverse(s1.begin(),s1.end());
                    s1.push_back(c[i]);
                    reverse(s1.begin(),s1.end());
                }else{
                    s2.push_back(c[i]);
                }
          }else{
              if(f[i] == 2){
                    reverse(s1.begin(),s1.end());
                    s1.push_back(c[i]);
                    reverse(s1.begin(),s1.end());
                }else{
                    s2.push_back(c[i]);
                }
          }
      }
  } 
  s = s1 + s2;
  if(count % 2 == 1){
      reverse(s.begin(),s.end());
  }
  cout<<s<<endl; 
}