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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void revStr(char* str){
  int size = strlen(str);
  int i,j;
  char tmp = {0};

  for(i = 0, j = size - 1; i < size / 2; i++, j--){
    tmp = str[i];
    str[i] = str[j];
    str[j] = tmp;
  }

  return;
}


int main(void){
  int q,t[200005],f[200005],count=0,muki=-1;
  char s[1000005],c[1000005];
  scanf("%s",s);
  scanf("%d",&q);
  for(int i=0;i<q;i++){
    scanf("%d",&t[i]);
    if(t[i]==2){
      scanf("%d %s",&f[i],c);
      if((muki==-1&&f[i]==1)||(muki==1&&f[i]==2)){
        strcat(c,s);
      }else if((muki==-1&&f[i]==2)||(muki==1&&f[i]==1)){
        strcat(s,c);
      }
    }else{
      muki=muki*(-1);
    }
  }
  if(muki==-1&&f[q-1]==1) printf("%s",c);
  if(muki==1&&f[q-1]==2){
    revStr(c);
    printf("%s",c);
  }
  if(muki==-1&&f[q-1]==2) printf("%s",s);
  if(muki==1&&f[q-1]==1) {
    revStr(s);
    printf("%s",s);
     }


  return 0;
}