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: Runtime Error)

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

int main() {
	char res[100000] = {};
	int q;
	deque<char> d;
	string s;
	cin >> s >> q;
	for(char& c: s) {
		d.push_back(c);
	}
	int rflag = 0;
	for(int i = 0; i < q; i++) {
		int t, f;
		char c;
		cin >> t;
		if (t == 1) {
			rflag ^= 1;
			continue;
		}
		cin >> f >> c;
		if (rflag == 0 && f == 1) {
			d.push_front(c);
		} else if (rflag == 0 && f == 2) {
			d.push_back(c);
		} else if (rflag == 1 && f == 1) {
			d.push_back(c);
		} else if (rflag == 1 && f == 2) {
			d.push_front(c);
		}
	}
	int i = 0;
	if (rflag) {
		for(auto di = d.rbegin(); di != d.rend(); ++di) {
			res[i++] = *di;
		}
	} else {
		for(auto di = d.begin(); di != d.end(); ++di) {
			res[i++] = *di;
		}
	}
	cout << res << endl;
}