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>
char s[10001];
void fanzhuan(int n)
{
	char s1[10001];
	int i,j=n-1;
	for(i=0;i<n;i++)
	{
		s1[i]=s[j];
		j--;
	}
	
	for(i=0;i<n;i++)
	{
		s[i]=s1[i];
	}
}
void tianjiatou(int n,char c)
{
	int i;
	char s1[10001];
	for(i=0;i<n;i++)
	{
		s1[i+1]=s[i];
	}
	s1[0]=c;
	for(i=0;i<n+1;i++)
	{
		s[i]=s1[i];
	}
}

int main()
{
	int i=0,j,n,k,a,b;
	char t[6];
	scanf("%s",s);
	while(s[i]!='\0') i++;
	n=i;
	scanf("%d ",&k);
	for(i=0;i<k;i++)
	{
		gets(t);
//		printf("%s\n",t);
		if(t[0]=='1') fanzhuan(n);
		else{
			if(t[2]=='1') tianjiatou(n,t[4]);
			else{
				s[n]=t[4];
			}
			n++;
		}
	}
	printf("%s",s);
	
	return 0;
}