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>
int main(void)
{
  char s[2][300001], c;
  int q, t, f, i, j, flg;

  scanf("%s", s[0]);                // 標準入力からの値取得
  scanf("%d", &q);                  // 標準入力からの値取得

  flg = 0;
  for (i = 0; i < q; i++) {
    scanf("%d %d %c", &t, &f, &c);  // 標準入力からの値取得
    if (t == 1) {
      for (i = 0; s[flg][i] != '\0'; i++)
        ;
      j = i;
      s[~flg][j--] = '\0';
      for (i = 0; s[flg][i] =!'\0'; i++) {
        s[~flg][j--] = s[flg][i];
      }
    } else {
      if (f == 1) {
        j = 0;
        s[~flg][j++] = c;
        for (i = 0; s[flg] != '\0'; i++) {
          s[~flg][j++] = s[flg][i];
        }
        s[~flg][j] = '\0';
      } else {
        for (i = 0, j = 0; s[~flg][i] != '\0'; i++, j++) {
          s[~flg][j] = s[flg][i];
        }
        s[~flg][j++] = c;
        s[~flg][j] = '\0';
      }
    }
    flg = ~flg;
    printf("%s\n", s[flg]);            // 出力
  }
  return 0;
}