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)

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<vector>
#include<functional>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
#include<cmath>
using namespace std;
typedef long long ll;
#define rep(i,a,b) for(int i=a;i<b;i++)
int main() {
	char str[50001];
	bool hanten = false;
	vector<char>ft, bk;
	int Q, T, F;
	char c[2];
	scanf("%s", str);
	scanf("%d", &Q);
	rep(i, 0, Q) {
		scanf("%d",&T);
		if (T == 1)hanten = !hanten;
		else {
			scanf("%d", &F);
			scanf("%s", c);
			getchar();
			(((F - 1) ^ hanten) ? bk : ft).push_back(c[0]);
		}
	}
	if (hanten) {
		int len = strlen(str);
		vector<char>str2(len + 1);
		str2[len] = 0;
		rep(i, 0, len) {
			str2[i] = str[len - i - 1];
		}
		reverse(bk.begin(), bk.end());
		bk.push_back(0);
		ft.push_back(0);
		printf("%s%s%s", bk.data(), str2.data(), ft.data());
	}
	else {
		reverse(ft.begin(), ft.end());
		ft.push_back(0);
		bk.push_back(0);
		printf("%s%s%s", ft.data(), str, bk.data());
	}
	return 0;
}