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: Accepted)

#include <bits/stdc++.h>
#define REP(i, n) for (int (i) = 0; (i) < (int)(n); i++) 
#define FOR(i, a, b) for(int (i) = a; (i) < (int)b; i++)
#define RREP(i, n) for(int (i)=((int)(n)-1); (i)>=0; i--)
#define RFOR(i, a, b) for(int (i) =((int)(b)-1); (i)>=(int)a; i--)
#define ALL(v) (v).begin(),(v).end()
#define MOD 1000000007
#define NIL -1
#define FI first
#define SE second
#define MP make_pair
#define PB push_back
#define SZ(x)  (int)x.size()
#define SP(x)  setprecision((int)x)


using namespace std ;
typedef long long ll;
typedef vector<int> vint;
typedef vector<vint> vvint;
typedef vector<string> vstr;
typedef pair<int, int> pii;
const int INF = 1e9;
const ll LINF = 1e18;
const double EPS = 1e-9;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;} //最大公約数
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} //最小公倍数

//-------------------------------------------------
//
void yes(){
  cout <<"Yes"<<endl ;
}
void no(){
  cout <<"No"<<endl ;
}


//-------------------------------------------------
//

//push_front() : void
//push_back() : void


//-------------------------------------------------


int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  string s ;
  cin >> s ;
  int q ;
  cin >> q ;
  int n = SZ(s) ;
  deque<char> data ;
  REP(i,n){
    data.push_back(s.at(i)) ;
  }
  
  int rev = 0 ;
  
  for(int i=0; i<q; i++){
    int qq ;
    cin >> qq ;
    if(qq==1){
      rev += 1 ;
    }
    else{
      int qqq ;
      char c ;
      cin >> qqq >> c ;
      if(rev%2==0){
        if(qqq==1){
        	data.push_front(c) ;
        }
      	else{
        	data.push_back(c) ;
      	}
      }
      else{
        if(qqq==2){
        	data.push_front(c) ;
        }
      	else{
        	data.push_back(c) ;
      	}    
      }
    }
  }
  
  int m = SZ(data) ;
  
  if(rev%2==1){
    RREP(i,m){
      cout << data.at(i) ;
    }
  }
  else{
     REP(i,m){
        cout << data.at(i) ;
     }
  }
  cout <<endl ;
  
  
  return 0 ;
}