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 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

char	*strrev(char *str)
{
	char *p1, *p2;

	if (!str || !*str)
		return (str);
	for (p1 = str, p2 = str + strlen(str) -1; p2 > p1; ++p1, --p2)
	{
		*p1 ^= *p2;
		*p2 ^= *p1;
		*p1 ^= *p2;
	}
	return (str);
}

char *add_back(char *str, char c)
{
	int index = 0;
	while (str[index] != '\0')
		index++;
	str[index] = c;
	str[index +1] = '\0';
	return str;
}

char *add_front(char *str, char c)
{
	int index = 0;
	while (str[index] != '\126')
		index++;
	str[index] = c;
	str[index -1] = '\126';
	return str;
}

int		main(void)
{
	char str[100001];
	char buf_str[500002];
	int q;

	scanf("%s", str);
	scanf("%d", &q);
	memset(buf_str, '0', 500000);
	buf_str[500001] = '\0';

	int index = 0;
	
	while (str[index] != '\0')
	{
		buf_str[230001 + index] = str[index];
		index++;
	}
	buf_str[230001+index] = '\0';
	buf_str[230000] = '\126';

	int dq;
	int f;
	char c;
	int flag = 1;

	index = 0;
	while (index <q)
	{
		scanf("%d", &dq);
		if (dq == 1)
		{
			flag = flag * (-1);
		}
		else if (dq == 2)
		{
			scanf("%d %c", &f, &c);
			if ((f == 1 && flag == 1) || (f == 2 && flag == -1))
				add_front(buf_str, c);
			else if ((f == 2 && flag ==1) || (f == 1 && flag == -1))
				add_back(buf_str, c);
		}
		index++;
	}
	index = 0;
	while (buf_str[index] != '\126')
		index++;
	char *r_str = &buf_str[index +1];
	if (flag == -1)
		strrev(r_str);
	printf("%s\n",r_str);
	return (0);
}