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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define rep(i,l,n) for(int i=l;i<n;i++)
#define ull unsigned long long
#define min(a,b) (a)<(b)?(a):(b)
#define max(a,b) (a)>(b)?(a):(b)

struct moji{
	char alp;
	struct moji *next;
	struct moji *back;
};

int main(){
	char s[100010];
	scanf("%s",s);
	struct moji *top = (struct moji*)malloc(sizeof(struct moji));
	struct moji *last = top;
	top->alp = s[0];
	top->next = NULL; top->back = NULL;
	for(int i=1;s[i]!=0;i++){
		last->next = (struct moji*)malloc(sizeof(struct moji));
		(last->next)->back = last;
		last = last->next;
		last->alp = s[i];
		last->next = NULL;
	}
	int Q;
	scanf("%d",&Q);
	rep(i,0,Q){
		int t;
		scanf("%d",&t);
		if( t==1 ){
			struct moji *temp=top;
			top=last; last=temp;
		}else{
			int f; char c;
			scanf("%d %c",&f,&c);
			struct moji *data;
			if( f==1 ) data=top;
			else data=last;
			if( data->next==NULL ){
				data->next = (struct moji*)malloc(sizeof(struct moji));
				(data->next)->back = data;
				data = data->next;
				data->alp = c;
				data->next = NULL;
				if(f==1) top = data;
				else last = data;
			}else{
				data->back = (struct moji*)malloc(sizeof(struct moji));
				(data->back)->next = data;
				data = data->back;
				data->alp = c;
				data->back = NULL;
				if(f==1) top = data;
				else last = data;
			}
		}
	}
	struct moji *data=top;
	if( top->next == NULL ){
		while( data!=NULL ){
			printf("%c",data->alp);
			data = data->back;
		}
	}else{
		while( data!=NULL ){
			printf("%c",data->alp);
			data = data->next;
		}
	}
	return 0;
}