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>
int main(void){
	char s[200001],rev[200001];
	int q,i,j,d,len,revo=0;
	scanf("%s%d",s,&q);
	int f[q+1];
	char c[q+1];
	for(i=0;i<q+1;i++){
		f[i]=0;
		c[i]='0';
	}
	for(i=0;i<q;i++){
		scanf("%d",&d);
		for(j=0;s[j]!='\0';j++);
		j--;;
		len=j;
		if(d==1){
			if(revo==0) revo=1;
			else revo=0;
		}else{
			scanf("%d %c",&f[i],&c[i]);
			if(revo==0){
				if(f[i]==1){
					for(j=0;j<len+2;j++) rev[j+1]=s[j];
					s[0]=c[i];
					for(j=1;j<len+3;j++) s[j]=rev[j];
				}else{
					s[len+1]=c[i];
					s[len+2]='\0';
				}
			}else{
				if(f[i]==0){
					for(j=0;j<len+2;j++) rev[j+1]=s[j];
					s[0]=c[i];
					for(j=1;j<len+3;j++) s[j]=rev[j];
				}else{
					s[len+1]=c[i];
					s[len+2]='\0';
				}
			}
		}
	}
	if(revo==1){
		while(j>=0){
				rev[len-j]=s[j];
				j--;
			}
			rev[len+1]='\0';
			for(j=0;j<len+1;j++) s[j]=rev[j];
	}
	printf("%s",s);
	return 0;
}