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>
#define ll unsigned long long
using namespace std;
string s;
int q;
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin>>s;
  int a,b;
  pair<string,int> l={"",0};
  pair<string ,int> r={"",0};
  cin>>q;
  while(q--)
  {
      cin>>a;
      if(a==1)
      {
          auto it=l;
          l.first=r.first;
          if(r.second==0)
          {
          l.second=r.first.size()-1;
          }
          else 
          {
               l.second=0;
          }


          r.first=it.first;
          if(it.second==0)
          {
               r.second=it.first.size()-1;
          }
          else
           r.second=0;
         
         
      }
      else
      {
          cin>>b;
          char ch;
          cin>>ch;
          if(b==1)
          {
            if(l.second!=0)
            {
               l.first+=ch;
               l.second+=1;
            }
            else
            {
                l.first.insert(l.first.begin(),ch);
            }
          }
          else 
          {
              if(r.second==0)
            {
               r.first+=ch;
            }
            else
            {
                r.first.insert(r.first.begin(),ch);
                r.second+=1;
            }
        
          }
      }
  } 
  string temp="";

  if(l.second==l.first.size()-1)
  {
    reverse(l.first.begin(),l.first.end());
    temp+=l.first;
  }
  else 
  {
      temp+=l.first;
  }
  temp+=s;
  if(r.second==r.first.size()-1)
  {
    reverse(r.first.begin(),r.first.end());
    temp+=r.first;
  }
  else 
  {
      temp+=r.first;
  }
  cout<<temp<<endl;
}