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>

void revstr(char* str){
  int size=strlen(str);
  int i,j;
  char tmp;
	
	for(i = 0,j = size - 1; i < size / 2; i++, j--){
		tmp = str[i];
		str[i] = str[j];
		str[j] = tmp;
	}
}

int main()
{
  int rev=0;
  int posi_judge;
  int do_judge;
  char main[100000];
  char pre[100000];
  char suf[100000];
  char tmp[2];
  int Q;
  int suf_cnt=0;
  int pre_cnt=0;
  
  scanf("%s",main);
  scanf("%d",&Q);
  for(int i=0;i<Q;i++)
  {
    scanf("%d",&do_judge);
    if (do_judge==1)
    {
      rev++;
    }
    else
    {
      scanf("%d %s",&posi_judge,tmp);
      if ((posi_judge==1 && rev%2==1)||(posi_judge==2 && rev%2==0))
        suf[suf_cnt++]=tmp[0];
      else
        pre[pre_cnt++]=tmp[0];
        
    }
}
  if(rev%2==0)
  {
    revstr(pre);
    printf("%s%s%s",pre,main,suf);
  }
  else
  {
    revstr(main);
    revstr(suf);
    printf("%s%s%s",suf,main,pre);
  }
}