Papyrus 的又一道谜题AI (CI翻译)
You are filled with determination.— Undertale
Papyrus came up with another puzzle for Frisk to solve. Papyrus brought two arrays $a$ and $b$ of length $n$ and allowed the following two operations to be performed:
Papyrus 又想出一道谜题让 Frisk 来解。Papyrus 给了 Frisk 两个长度为 $n$ 的数组 $a$ 和 $b$，并允许执行以下两种操作：AI (CI翻译)
- choose any index
$i$($1 \le i \le n$) and change$a_i$to$a_i - 1$. The execution time of such an operation is$1$second. - reorder all elements of array
$a$in any way. The execution time of such an operation is$c$seconds.
- 选择任意下标
$i$（$1 \le i \le n$），将$a_i$改为$a_i - 1$。执行该操作需要$1$秒。 - 以任意方式重新排列数组
$a$的所有元素。执行该操作需要$c$秒。
You need to convert array $a$ into array $b$.
你需要将数组 $a$ 转换为数组 $b$。AI (CI翻译)
Frisk wants to solve the puzzle as soon as possible. Help Frisk determine the minimum time needed to solve the puzzle. If there is no solution, output $-1$.
Frisk 想尽快解开谜题。请帮助 Frisk 求出解开谜题所需的最短时间。如果无解，输出 $-1$。AI (CI翻译)
Input
输入格式
AI (CI翻译)Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.
每个输入包含多组测试数据。第一行包含测试数据组数 $t$（$1 \le t \le 500$）。接下来是各组测试数据的描述。AI (CI翻译)
The first line of each test case contains two integers $n$ and $c$ ($1 \le n, c \le 100$) — the length of arrays $a$ and $b$ and the cost of the second operation.
每组测试数据的第一行包含两个整数 $n$ 和 $c$（$1 \le n, c \le 100$），分别表示数组 $a$ 和 $b$ 的长度，以及第二种操作的代价。AI (CI翻译)
The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the first array.
第二行包含 $n$ 个整数 $a_1, a_2, \ldots, a_n$（$1 \le a_i \le 100$），表示第一个数组的元素。AI (CI翻译)
The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 100$) — the elements of the second array.
第三行包含 $n$ 个整数 $b_1, b_2, \ldots, b_n$（$1 \le b_i \le 100$），表示第二个数组的元素。AI (CI翻译)
Output
输出格式
AI (CI翻译)For each test case, output a single integer representing the minimum number of seconds required to solve the puzzle, or $-1$ if it is impossible to solve the puzzle.
对于每组测试数据，输出一个整数，表示解开谜题所需的最少秒数。如果无解，则输出 $-1$。AI (CI翻译)
Note
说明
AI (CI翻译)In the first test case, it is impossible to transform $a$ into $b$ using only subtraction because $a_2 \lt b_2$. Let's rearrange the elements of array $a$ as follows: $[5, 2, 3] \Rightarrow [2, 3, 5]$. Now it is enough to subtract one from $a_3$, and we get that array $a$ becomes equal to array $b$ in $5 + 1 = 6$ seconds.
在第一组测试数据中，只通过减法操作无法将 $a$ 转换为 $b$，因为 $a_2 \lt b_2$。将数组 $a$ 的元素重新排列为 $[5, 2, 3] \Rightarrow [2, 3, 5]$。此时只需将 $a_3$ 减 $1$，数组 $a$ 就会与数组 $b$ 相等，总耗时为 $5 + 1 = 6$ 秒。AI (CI翻译)
In the second test case, all elements of $a$ are less than all elements of $b$, which means $a$ cannot be transformed into $b$.
在第二组测试数据中，$a$ 的所有元素都小于 $b$ 的所有元素，因此无法将 $a$ 转换为 $b$。AI (CI翻译)
In the third test case, you can choose not to rearrange the elements and get an answer of $3$. If you rearrange them at least once, the answer will be at least $4$, so the optimal answer is $3$ seconds.
在第三组测试数据中，可以不重新排列元素，得到答案 $3$。如果至少重新排列一次，答案至少为 $4$，因此最优答案是 $3$ 秒。AI (CI翻译)
In the sixth test case, the array can be rearranged as follows: $[14, 20, 20]$. It can be seen that the cost will then be $5 + (14 - 12) + (20 - 18) + (20 - 17) = 12$.
在第六组测试数据中，可以将数组重新排列为 $[14, 20, 20]$。此时总耗时为 $5 + (14 - 12) + (20 - 18) + (20 - 17) = 12$。
AI (CI翻译)
样例
6
3 5
5 2 3
2 3 4
3 3
1 2 3
4 5 6
4 4
4 5 2 3
3 5 1 2
6 4
2 4 5 3 6 8
5 8 3 1 2 5
5 11
5 8 11 14 17
16 12 10 10 6
3 5
20 14 20
12 18 17
6
-1
3
8
-1
12
题面与样例版权归 Codeforces 及出题人所有；中文翻译由 AI 生成（gpt-6.1-sol），如有出入请以原文为准。