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: Wrong Answer)

#define N 200002
#include <stdio.h>
#include <string.h>

char S[N];
char head[N];
long max;
void rev(long len)
{
	char temp;
	long cpymax = max-1;
	long j = 0;
	while (j < max)
	{
		temp = head[cpymax];
		head[cpymax] = S[len - 1 - j];
		S[len - 1 - j] = temp;
		cpymax--;
		j++;
	}
	j = len - j;
	for (long i = 0;i < ((j) / 2);i++)
	{
		temp = S[j-i-1];
		S[j-i-1] = S[i];
		S[i] = temp;
	}
}
int main()
{
	long len, Q;
	int type,mode;
	char word;
	fgets(S, N, stdin);
	len = strlen(S) - 1;
	S[len] = NULL;
	scanf("%d", &Q);
	for (long i = 0;i < Q;i++)
	{
		scanf("%d", &type);
		if (type == 1)
		{
			rev(len);
		}
		else
		{
			scanf("%d %c", &mode, &word);
			if (mode == 1)
			{
				head[max++] = word;
			}
			else
			{
				S[len++] = word;
			}
		}
	}
	for (long i = max - 1;i >= 0;i--)
	{
		printf("%c", head[i]);
	}
	printf(S);
	return 0;
}