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<stdlib.h>
#include<string.h>
#include<math.h>

#define MODSIZE 300005

char queue1[300005];
char queue2[300005];


int main(){
  int head1 = 0;
  int head2 = 0;
  int tail1 = 0;
  int tail2 = 0;
  char s[2];
  int i,q;
  int com1,com2;
  char t[2];
  int rev = 0;
  int l1,l2;

  scanf("%s", s);
  scanf("%d", &q);
  for(i = 0;i < q;i++){
    scanf("%d", &com1);

    if(com1 == 1) rev++;

    else{
      scanf("%d", &com2);
      scanf("%s", t);

      if(com2 == 1){
        if(rev % 2 == 0) queue1[head1++] = t[0];

        else queue2[head2++] = t[0];
      }

      else{
        if(rev % 2 == 0) queue2[head2++] = t[0];

        else queue1[head1++] = t[0];
      }
    }
  }

  l1 = strlen(queue1);
  l2 = strlen(queue2);

  if(rev % 2 == 0){
    head1--;
    while(head1 >= 0){
      printf("%c", queue1[head1]);
      head1--;
    }
    printf("%s%s\n", s, queue2);
  }

  else{
    head2--;
    while(head2 >= 0){
      printf("%c", queue2[head2]);
      head2--;
    }
    printf("%s%s\n", s, queue1);
  }
  
  return 0;
}