#AT2484. Ex - Bow Meow Optimization

Ex - Bow Meow Optimization

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Ex - Bow Meow Optimization

Score : $600$ points

Problem Statement

There are $N$ dogs numbered $1$ through $N$ and $M$ cats numbered $1$ through $M$. You will arrange the $(N+M)$ animals in a line from left to right. An arrangement causes each animal's frustration as follows:

  • The frustration of dog $i$ is $A_i\times|x-y|$, where $x$ and $y$ are the numbers of cats to the left and right of that dog, respectively.
  • The frustration of cat $i$ is $B_i\times|x-y|$, where $x$ and $y$ are the numbers of dogs to the left and right of that cat, respectively.

Find the minimum possible sum of the frustrations.

Constraints

  • $1\leq N,M \leq 300$
  • $1\leq A_i,B_i \leq 10^9$
  • All values in the input are integers.

Input

The input is given from Standard Input in the following format:

NN MM

A1A_1 A2A_2 \ldots ANA_N

B1B_1 B2B_2 \ldots BMB_M

Output

Print the answer as an integer.


2 2
1 3
2 4
6

Consider the following arrangement: dog $1$, cat $2$, dog $2$, cat $1$, from left to right. Then,

  • dog $1$'s frustration is $1\times|0-2|=2$;
  • dog $2$'s frustration is $3\times|1-1|=0$;
  • cat $1$'s frustration is $2\times|2-0|=4$; and
  • cat $2$'s frustration is $4\times|1-1|=0$,

so the sum of the frustrations is $6$. Rearranging the animals does not make the sum less than $6$, so the answer is $6$.


1 2
100
100 290
390

5 7
522 575 426 445 772
81 447 629 497 202 775 325
13354