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: Time Limit Exceeded)

#include <stdio.h>

int main(void){

  char s[300000+1];
  s[0] = '\n';
  int end = -1;
  for (int i = 0; i < 300000+1; i++) {
    scanf("%c", &s[i]);
    if (s[i] == '\n') {
      end = i-1;
      break;
    }
  }
  int q;
  scanf("%d", &q);
  int t[q][2];
  char tc[q];
  for (int i = 0; i < q; i++) {
    scanf("%d", &t[i][0]);
    if (t[i][0] == 2) {
      scanf("%d %c", &t[i][1], &tc[i]);
    } else {
      t[i][1] = 0;
      tc[i] = '0';
    }
  }

  char tmp;
  for (int i = 0; i < q; i++) {
    if (t[i][0] == 1) {
      for (int j = 0; j < end-j; j++) {
        tmp = s[j];
        s[j] = s[end-j];
        s[end-j] = tmp;
      }
    } else {
      end++;
      s[end+1] = s[end];
      if (t[i][1] == 1) {
        for (int j = end; j > 0; j--) {
          s[j] = s[j-1];
        }
        s[0] = tc[i];
      } else {
        s[end] = tc[i];
      }
    }
  }

  printf("%s", s);

  return 0;
}