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>
#include <math.h>
#include <cmath>

using namespace std;
using ll =long long;
using vi = vector<int>; 
using vvi = vector<vi>; 
using vl = vector<ll>; 
using vvl = vector<vl>; 

#define rep(i,n) ;for( int i =0; i < n ; i ++)

bool is_int_lround(double x){
  return std::lround(x)==x;
}
int ketasuu(int x){
  int n=0;
  while(x>0){
    x /=10 ;
    n ++;
  }
  return n;
}
ll gcd11(ll x, ll y){
  long long a= max(x,y), b=min(x,y), c=a%b;
  while( c!=0){
    a =b;
    b =c;
    c = a%b;
  }
  return b;
}

int abs(int x, int y){
  if(x>=y){
    return x-y;
  }
  else if(y>x){
    return y-x;
  }
}
int fiv(int n){
  if(n==0){
    return 0;
  }
  else if(n==1){
    return 1;
  }
  return fiv(n-1)+fiv(n-2);
}
ll conbi(int n,int m){
  
  cin>>n>>m;
  vector<ll> a(100);
  a[0] =1;
  for(int i=0;i<14;i++){
    a[i+1]=a[i]*(i+1);
  }
  return a[n] /(a[m] *a[n-m]);
}
int main(){
  string s;
  cin>>s;
  int q ;
  cin>> q;
  vector<tuple<int, int ,char>> sousa(q);
  rep(i,q){
    int res;
    cin >> res;
    if(res==1){
      sousa[i] = make_tuple(res,0,0);
    }
    else if(res ==2){
      int ress;
      char mozi;
      cin>>ress>>mozi;
      sousa[i]= make_tuple(res,ress,mozi);
    }
  }
  int numre=0;
  for(int i=0;i<q;i++){
    if(get<0>(sousa[i])==1){
      numre ++;
    }
    else{
      if(get<1>(sousa[i])==1 && numre %2==0){
        s = get<2>(sousa[i])+ s;
      }
      
      else if(get<1>(sousa[i])==1 && numre %2==1){
        s =  s+ get<2>(sousa[i]);
      }
      else if(get<1>(sousa[i]) ==2 && numre %2 ==0){
        s= s+get<2>(sousa[i]);
      }
      else if(get<1>(sousa[i]) ==2 &&numre%2 ==1){
        s = get<2>(sousa[i]) +s ;
      }
    }
  }
  if(numre %2 ==1){
    rep(i,s.size()){
      cout <<s[s.size()-i];
    }
    cout<<" "<<endl;
  }
  else{
    cout <<s<<endl;
  }
}