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>
using namespace std;

template<typename T, typename U> bool chmin(T &a, const U &b) {
	return (a > b ? a = b, true : false);
}
template<typename T, typename U> bool chmax(T &a, const U &b) {
	return (a < b ? a = b, true : false);
}

typedef long long ll;

#define writeln(n) cout<<n<<"\n";

typedef pair<int, int> P;
typedef pair<string, int> Psi;

ll numct[210000];
ll k[210000];
ll cnb[210000];
int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	string s;
	int q;
	cin>>s>>q;

	bool dire = true;
	for(int i=0; i<q; i++){
		int t,f;string ss;
		cin>>t;

		if(t ==1){
			if(dire){
				dire = false;
			}else {
				dire =true;
			}
		}else if(dire){
			cin>>f>>ss;
			if(f ==1){
				s.insert(0,ss);
			}else {
				s +=ss;
			}
		}else {
			cin>>f>>ss;
			if(f ==1){
				s +=ss;
			}else {
				s.insert(0,ss);
			}
		}

/*		cout<<s<<endl;*/
	}

	if(!dire){
		reverse(s.begin(),s.end());
	}

	cout<<s<<endl;
}