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: Runtime Error)

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

int zengo(char *s,long N){
	char s_1[100100];
	long i;
	for(i=N-1;i>=0;i--){
		s_1[i]=s[N-1-i];
	}
	
	strcpy(s,s_1);
	return 0;
}

int main(){
	int A,B;
	long len,N,i,j;
	char s[100100],c;
	
	scanf("%s",s);
	len = strlen(s);
	
	scanf("%ld",&N);
	
	for(i=0;i<N;i++){
		scanf("%d",&A);
		
		//printf("(%d:%d)\n",i,A);
		if(A==1){
			zengo(s,len);
		}
		else if(A==2){
			scanf("%d %c",&B,&c);
		//	printf("%d %c\n",B,c);
			len++;
			if(B==1){
				for(j=len-1;j>0;j--){
					s[j]=s[j-1];
				}
				s[0]=c;
			}
			else{
				
				s[len-1]=c;
			}
		}
	}
	//s[len]='\0';
	//for(i=0;i<len;i++){
	//	printf("%c",s[i]);
	//}
		printf("%s\n",s);
	
 return 0;
 }