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>
int main(){
  char s[100001];
  int q,i,rev=0,len;
  scanf("%s",s);
  scanf("%d",&q);
  len=strlen(s);
  char be[q+1],af[q+1];
  int benum=0,afnum=0;
  int t[q+1],f;
  for(i=0;i<q;i++){
    scanf("%d",&t[i]);
    if(t[i]==1)
      rev++;
    else{
      printf("%d\n\n",i);
      scanf("%d ",&f);
      if((rev+f)%2==0)
        scanf("%c",&af[afnum++]);
      else
        scanf("%c",&be[benum++]);
    }
  }
  if(rev%2==0){
    for(i=benum-1;i>=0;i--)
      printf("%c",be[i]);
    for(i=0;i<len;i++)
      printf("%c",s[i]);
    for(i=0;i<afnum;i++)
      printf("%c",af[i]);
    }
  else{
    for(i=afnum-1;i>=0;i--)
      printf("%c",af[i]);
    for(i=len-1;i>=0;i--)
      printf("%c",s[i]);
    for(i=0;i<benum;i++)
      printf("%c",be[i]);
    }
  printf("\n");
  return 0;
}