10872: ABC158 - D - String Formation
Description
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.
Input
```
$S$
$Q$
$Query_1$
$:$
$Query_Q$
```
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
Constraints
- $S$ 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.
Sample 1 Input
a
4
2 1 p
1
2 2 c
1
Sample 1 Output
cpa
There will be $Q = 4$ operations. Initially, $S$ is a
.
-
Operation $1$: Add
p
at the beginning of $S$. $S$ becomespa
. -
Operation $2$: Reverse $S$. $S$ becomes
ap
. -
Operation $3$: Add
c
at the end of $S$. $S$ becomesapc
. -
Operation $4$: Reverse $S$. $S$ becomes
cpa
.
Thus, the resulting string is cpa
.
Sample 2 Input
a
6
2 2 a
2 1 b
1
2 2 c
1
1
Sample 2 Output
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 3 Input
y
1
2 1 x
Sample 3 Output
xy