AB Balance
Description
You are given a string $s$ of length $n$ consisting of characters a and/or b.
Let $\operatorname{AB}(s)$ be the number of occurrences of string ab in $s$ as a substring. Analogically, $\operatorname{BA}(s)$ is the number of occurrences of ba in $s$ as a substring.
In one step, you can choose any index $i$ and replace $s_i$ with character a or b.
What is the minimum number of steps you need to make to achieve $\operatorname{AB}(s) = \operatorname{BA}(s)$?
Reminder:
The number of occurrences of string $d$ in $s$ as substring is the number of indices $i$ ($1 \le i \le |s| - |d| + 1$) such that substring $s_i s_{i + 1} \dots s_{i + |d| - 1}$ is equal to $d$. For example, $\operatorname{AB}($aabbbabaa$) = 2$ since there are two indices $i$: $i = 2$ where aabbbabaa and $i = 6$ where aabbbabaa.
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.
The first and only line of each test case contains a single string $s$ ($1 \le |s| \le 100$, where $|s|$ is the length of the string $s$), consisting only of characters a and/or b.
For each test case, print the resulting string $s$ with $\operatorname{AB}(s) = \operatorname{BA}(s)$ you'll get making the minimum number of steps.
If there are multiple answers, print any of them.
Input
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.
The first and only line of each test case contains a single string $s$ ($1 \le |s| \le 100$, where $|s|$ is the length of the string $s$), consisting only of characters a and/or b.
Output
For each test case, print the resulting string $s$ with $\operatorname{AB}(s) = \operatorname{BA}(s)$ you'll get making the minimum number of steps.
If there are multiple answers, print any of them.
Samples
4
b
aabbbabaa
abbb
abbaab
b
aabbbabaa
bbbb
abbaaa
Note
In the first test case, both $\operatorname{AB}(s) = 0$ and $\operatorname{BA}(s) = 0$ (there are no occurrences of ab (ba) in b), so can leave $s$ untouched.
In the second test case, $\operatorname{AB}(s) = 2$ and $\operatorname{BA}(s) = 2$, so you can leave $s$ untouched.
In the third test case, $\operatorname{AB}(s) = 1$ and $\operatorname{BA}(s) = 0$. For example, we can change $s_1$ to b and make both values zero.
In the fourth test case, $\operatorname{AB}(s) = 2$ and $\operatorname{BA}(s) = 1$. For example, we can change $s_6$ to a and make both values equal to $1$.