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: Accepted)

/**
*    author:  yuya1234
*    created: 13.07.2020 16:54:10
**/

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define REPD(i,n) for(ll i=n-1;i>=0;i--)
#define FOR(i,a,b) for(ll i=a;i<=(ll)(b);i++)
#define FORD(i,a,b) for(ll i=a;i>=(ll)(b);i--)
#define FORV(i,v) for(auto i=v.begin(); i!=v.end();i++)

#define SORT(s) sort((s).begin(),(s).end())
#define SORTD(s) sort((s).rbegin(),(s).rend())
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );

#define SZ(x) ((int)(x).size())
#define MEMSET(v, h) memset((v), h, sizeof(v))

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
int gcd(int a,int b){return b?gcd(b,a%b):a;}

int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  //std::cout << std::fixed << std::setprecision(15);

  list<string> s;
  string tmp;
  cin>>tmp;
  s.push_back(tmp);

  bool rev=false;
  int n;
  cin>>n;
  REP(i,n){
    int q;
    cin>>q;
    if(q==1){
      rev=!rev;
    }
    else{
      int f;
      string c;
      cin>>f>>c;
      if((!rev && f==1) || (rev && f==2)){
        s.emplace_front(c);
      }
      else{
        s.emplace_back(c);
      }
    }
  }

  if(rev){
    for(auto itr = s.rbegin(); itr != s.rend(); ++itr) {
      reverse((*itr).begin(),(*itr).end());
      cout<<*itr;
    }
  }
  else{
    for(auto itr = s.begin(); itr != s.end(); ++itr) {
      cout<<*itr;
    }
  }
  
  cout<<endl;
  

  return 0;
}