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: Accepted)

#include <stdio.h>
int main(void)
{
  char s[500001], c, ans[300001];
  int q, t, f, i, j, left_end, right_end, rightward_flg;

  scanf("%s", &s[200000]);            // 標準入力からの値取得
  scanf("%d", &q);                    // 標準入力からの値取得

  rightward_flg = 1;
  left_end = 199999;
  for (right_end = 200000; s[right_end] != '\0'; right_end++)
    ;
  
  for (i = 0; i < q; i++) {
    scanf("%d", &t);                  // 標準入力からの値取得
    if (t == 1) {
      rightward_flg = ~rightward_flg;
    } else {
      scanf("%d %c", &f, &c);         // 標準入力からの値取得
      if (f == 1) {
        if (rightward_flg == 1) {
          s[left_end--] = c;
        } else {
          s[right_end++] = c;
        }
      } else {
        if (rightward_flg == 1) {
          s[right_end++] = c;
        } else {
          s[left_end--] = c;
        }
      }
    }
  }

  // 出力
  if (rightward_flg == 1) {
    s[right_end] = '\0';
    printf("%s\n", &s[++left_end]);
  } else {
    for (i = --right_end, j = 0; i > left_end; i--) {
      ans[j++] = s[i];
    }
    ans[j] = '\0';
    printf("%s\n", ans);
  }
  return 0;
}