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)

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

int main(void)
{

	char c_str[100000];
	char r_str[200000];
	char l_str[200000];
	char ans[300000];
	
	unsigned long dir = 0;
	unsigned long num;
	unsigned long num_q2;

	char que[3];
	int i=0,j,k,l;
	char que2[1];
	
	
	unsigned long c_len,r_len,l_len;
	
	scanf("%s",c_str);
	c_len = strlen(c_str);
	
	scanf("%lu",&num);

	num_q2 = num;
	


	while(i < num_q2){
		i++;
		scanf("%s",que);
		num_q2 = num;
		
		que2[0] = que[2];
		if(que[0] == '1'){
			++dir;
		}else{
			if(dir%2 == 0){
				if(que[1] == '1'){
					strcat(l_str,que2);
				}else{
					strcat(r_str,que2);
				}
			}else{
				if(que[1] == '1'){
					strcat(r_str,que2);
				}else{
					strcat(l_str,que2);
				}
			}
		}
	}
	
	c_len = strlen(c_str);
	r_len = strlen(r_str);
	l_len = strlen(l_str);
	
	if(dir%2 == 0){
		for(j=0;j<l_len;j++)
			ans[j]=l_str[l_len-1-j];
		for(k=0;k<c_len;k++)
			ans[k+l_len] = c_str[k];
		for(l=0;l<r_len;l++)
			ans[l+l_len+c_len] = r_str[l];
		printf("%s\n",ans);
	}else{
		for(j=0;j<r_len;j++)
			ans[j]=r_str[r_len-1-j];
		for(k=0;k<c_len;k++)
			ans[k+r_len]=c_str[c_len-1-k];
		for(l=0;l<l_len;l++)
			ans[l+r_len+c_len]=l_str[l];
		printf("%s\n",ans);
	}
	
	return 0;
}