AB Balance

### 题意简述 给定一个字符串 $s$，定义 $\mathrm{AB}(s)$ 为 $s$ 中串 `ab` 出现的次数，$\mathrm{BA}(s)$，为 `ba` 出现的次数。 每次可以修改一个字符，要求通过尽量少的操作 ，使得 $\mathrm{AB}(s)=\mathrm{BA}(s)$。输出修改后的字符串。 ### 输入格式 多组测试，先输入整数 $T\ (1\le T\le 1000)$。 接下来 $T$ 行，每行一个字符串 $s\ (1\le|s|\le 100)$。 ### 输出格式 对于每组数据，输出一个字符串表示答案。

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.

4
b
aabbbabaa
abbb
abbaab

b
aabbbabaa
bbbb
abbaaa

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 $ .