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 <iostream>
#include <cmath>
#include <climits>
#include <algorithm>
#include <vector>
#include <unordered_map>
#include <unordered_set>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <string.h>
#include <iomanip>
#include <ctype.h>

#define ll          long long int
#define ull         unsigned long long int
#define pb          push_back
#define pii         pair<int,int>
#define F           first
#define S           second
#define vi          vector<int>
#define vli          vector<ll>
#define vii         vector<pii>
#define mi          map<int,int>
#define mii         map<pii,int>
#define M_PI        3.14159265358979323846
#define mp          make_pair
#define sz(x)       (int)x.size()
//#define hell        1000000007
#define matrix      vector<vector<ll>>
#define forn(i,a,b)  for(int i=a;i<b;i++)
#define eps 0.0000000001
#define all(a)      (a).begin(),(a).end()
const long long INF = (long long) 1e7+1;
const long long mod = (long long) 998244353 ;
const long long hell = (long long) 1000000007;

using namespace std;
#include <stdio.h>

ll power(ll x,ll y)
{
  ll res=1;
  while(y>0){
    if(y%2) res=(x%hell*res%hell)%hell;
    x=(x%hell*x%hell)%hell;
    y=y/2;
  }
  return res;
}

ll modinv(ll x){
  return power(x,hell-2);
}

void solve();


void fastIO() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}

int main(){
   fastIO();
   int test;
   //cin>>test;
   test=1;
   
   while(test--)
   solve();
}
//void sieve();
ll pw(ll k,ll x);
ll gcd(ll x,ll y);

/*
  For freq or prime factors

for(auto pri:prime){
  while(k%pri==0){
    pcnt[pri]++;
    k/=pri;
  }
}*/

void solve(){
    string s;
    cin>>s;
    
    int q;
    cin>>q;
    int cur=0;
    while(q--){
        int t;
        cin>>t;
        
        if(t==1){
            cur++;
        }
        else{
            int f;
            cin>>f;
            string c;
            cin>>c;
            
            if(cur&1){
               if(f==1)
                s=s+c;
                else
                s=c+s;
            }
            else{
                if(f==1)
                s=c+s;
                else
                s=s+c;
            }
        }
    }
    if(cur&1)
    reverse(s.begin(),s.end());
    cout<<s;
}    


///Before submit=>
///    *check for integer overflow,array bounds
///    *check for n=1

ll gcd(ll a,ll b)
{
    while( b != 0)
	{
		ll  t = b;
		b= a %t;
		a = t;
	}
	return a;
}


ll pw(ll k,ll x){
  if(x==0)
  return 1;
  if(x&1)
  return ((pw(k,x-1))*k);
  else
  return ((pw(k,x/2))*(pw(k,x/2)));
}