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<bits/stdc++.h>
using namespace std;

#define ll long long
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define endl "\n"
#define pb push_back
#define all(a) (a).begin(),(a).end()
#define rev(a) (a).rbegin(),(a).rend()
#define pll pair<ll,ll>
#define rep(i,a,b) for(long long i = (a); i <= (b); i++)

const int mod = 1e9+7;
const int N = 1e5+5; 

int main(){
	fast;
	ll t = 1; 
	//cin >> t;
	while(t--){
	   string s; cin >> s;
	   ll q; cin >> q;
	   bool normal = true;
	   while(q--){
	       ll q1; cin >> q1;
	       if(q1 == 1){
	           if(normal) normal = false;
	           else normal = true;
	       }
	       else{
	           ll f; cin >> f;
	           char ch; cin >> ch;
	           if(normal){
	               if(f == 1) s = ch + s;
	               else s = s + ch;
	           }
	           else{
	               if(f == 1) s = s + ch;
	               else s = ch + s;
	           }
	       }
	   }
	   
	   if(normal) cout << s << endl;
	   else{
	       reverse(all(s));
	       cout << s << endl;
	   }
	}
}