CF1798F.Gifts from Grandfather Ahmed

Grandfather Ahmed's School has $n+1$ students. The students are divided into $k$ classes, and $s_i$ students study in the $i$-th class. So, $s_1 + s_2 + \ldots + s_k = n+1$.

Due to the upcoming April Fools' Day, all students will receive gifts!

Grandfather Ahmed planned to order $n+1$ boxes of gifts. Each box can contain one or more gifts. He plans to distribute the boxes between classes so that the following conditions are satisfied:

1. Class number $i$ receives exactly $s_i$ boxes (so that each student can open exactly one box).
2. The total number of gifts in the boxes received by the $i$-th class should be a multiple of $s_i$ (it should be possible to equally distribute the gifts among the $s_i$ students of this class).

Unfortunately, Grandfather Ahmed ordered only $n$ boxes with gifts, the $i$-th of which contains $a_i$ gifts.

Ahmed has to buy the missing gift box, and the number of gifts in the box should be an integer between $1$ and $10^6$. Help Ahmed to determine, how many gifts should the missing box contain, and build a suitable distribution of boxes to classes, or report that this is impossible.

艾哈迈德爷爷的学校共有 $n+1$ 名学生。这些学生被分入 $k$ 个班级，其中第 $i$ 个班级有 $s_i$ 名学生。因此，满足 $s_1 + s_2 + \ldots + s_k = n+1$。

由于即将到来的愚人节，所有学生都将收到礼物！

艾哈迈德爷爷原计划订购 $n+1$ 个礼物盒。每个盒子可装一个或多个礼物。他打算将这些盒子分配给各个班级，使得满足以下条件：

1. 第 $i$ 个班级恰好收到 $s_i$ 个盒子（从而保证每名学生恰好打开一个盒子）；
2. 第 $i$ 个班级所收到的所有盒子中的礼物总数必须是 $s_i$ 的倍数（即这些礼物可以被该班的 $s_i$ 名学生平均分配）。

不幸的是，艾哈迈德爷爷只订购了 $n$ 个装有礼物的盒子，其中第 $i$ 个盒子含有 $a_i$ 个礼物。

艾哈迈德必须购买缺失的那个礼物盒，且该盒中礼物数量必须为介于 $1$ 到 $10^6$ 之间的整数。请帮助艾哈迈德确定：缺失的盒子中应装多少个礼物，并构造一种满足上述条件的盒子到班级的分配方案；若不可能实现，请报告这一点。

The first line of the input contains two integers $n$ and $k$ ($1 \le n, k \le 200$, $k \le n + 1$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the number of gifts in the available boxes.

The third line contains $k$ integers $s_1, s_2, \ldots, s_k$ ($1 \le s_i \le n+1$) — the number of students in classes. It is guaranteed that $\sum s_i = n+1$.

输入的第一行包含两个整数 $n$ 和 $k$（$1 \le n, k \le 200$，且 $k \le n + 1$）。

第二行包含 $n$ 个整数 $a_1, a_2, \ldots, a_n$（$1 \le a_i \le 10^6$）——表示可用盒子中礼物的数量。

第三行包含 $k$ 个整数 $s_1, s_2, \ldots, s_k$（$1 \le s_i \le n+1$）——表示各班级的学生人数。保证 $\sum s_i = n+1$。

If there is no way to buy the remaining box, output the integer $-1$ in a single line.

Otherwise, in the first line, output a single integer $s$ — the number of gifts in the box that Grandfather Ahmed should buy ($1 \le s \le 10^6$).

Next, in $k$ lines, print the distribution of boxes to classes. In the $i$-th line print $s_i$ integers — the sizes of the boxes that should be sent to the $i$-th class.

If there are multiple solutions, print any of them.

如果无法购买剩余的盒子，则在单独一行中输出整数 $-1$。

否则，在第一行输出一个整数 $s$ —— 祖父艾哈迈德应购买的盒子中的礼物数量（$1 \le s \le 10^6$）。

接下来，在 $k$ 行中，输出盒子分配给各班级的方案。在第 $i$ 行输出 $s_i$ 个整数 —— 应发送给第 $i$ 个班级的盒子的尺寸。

若存在多种解法，输出任意一种即可。

In the first test, Grandfather Ahmed can buy a box with just $1$ gift. After that, two boxes with $7$ gifts are sent to the first class. $7 + 7 = 14$ is divisible by $2$. And the second class gets boxes with $1, 7, 127$ gifts. $1 + 7 + 127 = 135$ is evenly divisible by $3$.

In the second test, the classes have sizes $6$, $1$, $9$, and $3$. We show that the available boxes are enough to distribute into classes with sizes $6$, $9$, $3$, and in the class with size $1$, you can buy a box of any size. In class with size $6$ we send boxes with sizes $7$, $1$, $7$, $6$, $5$, $4$. $7 + 1 + 7 + 6 + 5 + 4 = 30$ is divisible by $6$. In class with size $9$ we send boxes with sizes $1$, $2$, $3$, $8$, $3$, $2$, $9$, $8$, $9$. $1 + 2 + 3 + 8 + 3 + 2 + 9 + 8 + 9 = 45$ is divisible by $9$. The remaining boxes ($6$, $5$, $4$) are sent to the class with size $3$. $6 + 5 + 4 = 15$ is divisible by $3$.

在第一个测试用例中，祖父艾哈迈德可以购买一个仅含 $1$ 件礼物的盒子。随后，向第一个班级发送两个各含 $7$ 件礼物的盒子，$7 + 7 = 14$ 可被 $2$ 整除。第二个班级则收到含 $1$、$7$、$127$ 件礼物的盒子，$1 + 7 + 127 = 135$ 可被 $3$ 整除。

在第二个测试用例中，各班级人数分别为 $6$、$1$、$9$ 和 $3$。我们说明：现有盒子足以分配给大小为 $6$、$9$、$3$ 的班级；而对于人数为 $1$ 的班级，可购买任意大小的盒子。在人数为 $6$ 的班级中，我们发送大小分别为 $7$、$1$、$7$、$6$、$5$、$4$ 的盒子，其总和 $7 + 1 + 7 + 6 + 5 + 4 = 30$ 可被 $6$ 整除。在人数为 $9$ 的班级中，我们发送大小分别为 $1$、$2$、$3$、$8$、$3$、$2$、$9$、$8$、$9$ 的盒子，其总和 $1 + 2 + 3 + 8 + 3 + 2 + 9 + 8 + 9 = 45$ 可被 $9$ 整除。剩余的盒子（大小为 $6$、$5$、$4$）发送至人数为 $3$ 的班级，其总和 $6 + 5 + 4 = 15$ 可被 $3$ 整除。