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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

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

int main(){
	ios::sync_with_stdio(false);
	string s, c;
	int n, t, f;
	vector<string> vet[2];
	cin >> s;
	cin >> n;
	bool flag = true;
	while(n--){
		cin >> t;
		if (t == 1)
			flag = !flag;
		else{
			cin >> f;
			cin >> c;
			if (f == 1 && flag)
				vet[0].push_back(c);
			else if (f == 1 && !flag)
				vet[1].push_back(c);
			else if (f == 2 && flag)
				vet[1].push_back(c);
			else
				vet[0].push_back(c);
		}
	}
	if (flag){
		for (int i = vet[0].size() - 1; i >= 0; i--)
			cout << vet[0][i];
		cout << s;
		for (int i = 0; i < vet[1].size(); i++)
			cout << vet[1][i];
	}
	else{
		for (int i = vet[1].size() - 1; i >= 0; i--)
			cout << vet[1][i];
		cout << s;
		for (int i = 0; i < vet[0].size(); i++)
			cout << vet[0][i];
	}
	cout << endl;
	return 0;
}