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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>

void inverse(char *s, int len)
{
  char temp;
  int i, n;
  
  n = len / 2;
  
  for (i=0; i<n; i++) {
    temp = s[i];
    s[i] = s[len-i-1];
    s[len-i-1] = temp;
  }
  
  return;
}

void append(char *s, char c, int len, int type)
{
  int i;
  
  switch (type) {
    case 1:
      for (i=len; i>0; i--) {
        s[i] = s[i-1];
      }
      s[0] = c;
      break;
      
    case 2:
      s[len] = c;
      break;
  }
  
  s[len+1] = '\n';

  return;
}

int main (void)
{
  char s[500000];
  int q, t, f, i, len;
  char c;
  
  scanf("%s", s);
  len = strlen(s);
  
  scanf("%d", &q);
  
  for (; q>0; q--) {
    scanf ("%d", &t);
    
    if (t == 1) {
      inverse(s, len);
    } else {
      scanf("%d %c", &f, &c);
      append(s, c, len, f);
      len++;
    }
  }
  
  printf("%s", s);
  
  return 0;
}