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: Accepted)

#include<iostream>
#include<vector>
#include<deque>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)
using ll = long long;
using P = pair<int, int>;
#include<algorithm>
#include<math.h>
#include<map>
#include<queue>
#include<set>

int main() {
	string s;
	int Q;
	cin >> s>>Q;
	deque<char> q;
	rep(i, s.size())q.push_back(s[i]);
	//cout << q.front() << endl;
	int sw(1);
	rep(i, Q) {
		int buf;
		cin >> buf;
		if (buf == 1)sw *= -1;
		if (buf == 2) {
			int f;
			char c;
			cin >> f >> c;
			if (f == 1) {
				if (sw == 1)q.push_front(c);
				else q.push_back(c);
			}
			else {
				if (sw == -1)q.push_front(c);
				else q.push_back(c);
			}
		}
	}
	int qs = q.size();
	rep(i, qs) {
		
		if (sw == 1) {
			cout << q.front(); q.pop_front();
		}
		else {
			cout << q.back(); q.pop_back();
		}
		//cout << endl;
	}
	return 0;
}