CF1363F.Rotating Substrings

You are given two strings $s$ and $t$, each of length $n$ and consisting of lowercase Latin alphabets. You want to make $s$ equal to $t$.

You can perform the following operation on $s$ any number of times to achieve it —

• Choose any substring of $s$ and rotate it clockwise once, that is, if the selected substring is $s[l,l+1...r]$, then it becomes $s[r,l,l + 1 ... r - 1]$. All the remaining characters of $s$ stay in their position.

  For example, on rotating the substring $[2,4]$ , string "abcde" becomes "adbce".

A string $a$ is a substring of a string $b$ if $a$ can be obtained from $b$ by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

Find the minimum number of operations required to convert $s$ to $t$, or determine that it's impossible.

给你两个长度均为 $n$ 的字符串 $s$ 和 $t$，它们均由小写拉丁字母组成。你想让 $s$ 变为与 $t$ 相等。

你可以对 $s$ 执行以下操作任意多次，以达成目标：

• 任选 $s$ 的一个子串，并将其顺时针旋转一次：即若所选子串为 $s[l,l+1\ldots r]$，则它变为 $s[r,l,l + 1 \ldots r - 1]$；$s$ 中其余字符位置保持不变。
  例如，将子串 $[2,4]$（下标从 1 开始）旋转后，字符串 "abcde" 变为 "adbce"。

字符串 $a$ 是字符串 $b$ 的子串，当且仅当 $a$ 可通过从 $b$ 的开头删除若干（可能为零或全部）字符、并从 $b$ 的末尾删除若干（可能为零或全部）字符而得到。

求将 $s$ 转换为 $t$ 所需的最少操作次数；若不可能实现，则判定为不可能。

The first line of the input contains a single integer $t$ $(1\leq t \leq 2000)$ — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ $(1\leq n \leq 2000)$ — the length of the strings.

The second and the third lines contain strings $s$ and $t$ respectively.

The sum of $n$ over all the test cases does not exceed $2000$.

输入的第一行包含一个整数 $t$（$1\leq t \leq 2000$）——测试用例的数量。随后是各测试用例的描述。

每个测试用例的第一行包含一个整数 $n$（$1\leq n \leq 2000$）——字符串的长度。

第二行和第三行分别包含字符串 $s$ 和 $t$。

所有测试用例的 $n$ 值之和不超过 $2000$。

For each test case, output the minimum number of operations to convert $s$ to $t$. If it is not possible to convert $s$ to $t$, output $-1$ instead.

对于每个测试用例，输出将 $s$ 转换为 $t$ 所需的最少操作次数。如果无法将 $s$ 转换为 $t$，则输出 $-1$。

For the $1$-st test case, since $s$ and $t$ are equal, you don't need to apply any operation.

For the $2$-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.

For the $3$-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.

For the $4$-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length $2$ beginning at the second character to convert it to cba.

For the $5$-th test case, you only need to apply one operation on the entire string abab to convert it to baba.

For the $6$-th test case, it is not possible to convert string $s$ to $t$.

对于第 $1$ 个测试用例，由于 $s$ 和 $t$ 相等，因此无需执行任何操作。

对于第 $2$ 个测试用例，只需对整个字符串 ab 执行一次操作，即可将其转换为 ba。

对于第 $3$ 个测试用例，只需对整个字符串 abc 执行一次操作，即可将其转换为 cab。

对于第 $4$ 个测试用例，需要执行两次操作：首先对整个字符串 abc 执行操作，将其转换为 cab；然后对从第二个字符开始、长度为 $2$ 的子串执行操作，将其转换为 cba。

对于第 $5$ 个测试用例，只需对整个字符串 abab 执行一次操作，即可将其转换为 baba。

对于第 $6$ 个测试用例，无法将字符串 $s$ 转换为 $t$。