[Codeforces 500C] New Year Book Reading (贪心?)

题目传送门：点击传送

Time limit per test：2 seconds

Memory limit per test：256 megabytes

Description

New Year is coming, and Jaehyun decided to read many books during 2015, unlike this year. He has n books numbered by integers from 1 to n. The weight of the i-th (1 ≤ i ≤ n) book is wi.

As Jaehyun's house is not large enough to have a bookshelf, he keeps the n books by stacking them vertically. When he wants to read a certain book x, he follows the steps described below.

1. He lifts all the books above book x.
2. He pushes book x out of the stack.
3. He puts down the lifted books without changing their order.
4. After reading book x, he puts book x on the top of the stack.

He decided to read books for m days. In the j-th (1 ≤ j ≤ m) day, he will read the book that is numbered with integer bj (1 ≤ bj ≤ n). To read the book, he has to use the process described in the paragraph above. It is possible that he decides to re-read the same book several times.

After making this plan, he realized that the total weight of books he should lift during m days would be too heavy. So, he decided to change the order of the stacked books before the New Year comes, and minimize the total weight. You may assume that books can be stacked in any possible order. Note that book that he is going to read on certain step isn't considered as lifted on that step. （拿要看的那本书不算lift）Can you help him?

Input

The first line contains two space-separated integers n (2 ≤ n ≤ 500) and m (1 ≤ m ≤ 1000) — the number of books, and the number of days for which Jaehyun would read books.

The second line contains n space-separated integers w1, w2, ..., wn (1 ≤ wi ≤ 100) — the weight of each book.

The third line contains m space separated integers b1, b2, ..., bm (1 ≤ bj ≤ n) — the order of books that he would read.

Note that he can read the same book more than once.（每本书可看多次）

Output

Print the minimum total weight of books he should lift, which can be achieved by rearranging the order of stacked books.

Examples

input

3 5
1 2 3
1 3 2 3 1

output

12

Note

Here's a picture depicting the example. Each vertical column presents the stacked books.

 

解析

题意：给n本书你，每本书对应一个重量。要看m天的书，当天要看的书需要放在表面。如果当天要看的书被压着，就要抽起上面的书（lift），最后算总共lift的最小值。（具体过程可看Hint图片）

思路：想要lift最小，自然就是把要看的书放在前面。按照看书的顺序（第一次出现）排好顺序，然后模拟过程，得到的即为答案。

 

代码

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
int w[505];
int day[1005];
int a[505];
int stat[505];
int main()
{
	int n,m,i,j,num=1,ans=0;
	while(~scanf("%d%d",&n,&m))
	{
		memset(stat,0,sizeof(stat));
		for(i=1;i<=n;i++)
		{
			scanf("%d",&w[i]);//i从1开始，w[i]表示第i的书的重量
		}
		for(i=1;i<=m;i++)
		{
			scanf("%d",&day[i]);//day[i]表示第i天读的书的编号
			if(stat[day[i]]==0)
			{
				a[num]=day[i];
				num++;
				stat[day[i]]=1;
			}
		}
		for(i=1;i<=m;i++)
		{
			int pos=1;
			while(a[pos]!=day[i])//把要拿的那本书上面的重量都加上
			{
				ans=ans+w[a[pos]];
				pos++;
			}
			for(j=pos;j>0;j--)//把上面的书都下移一位
			{
				a[j]=a[j-1];
			}
			a[1]=day[i];//让最上面的书为当天读的书
		}
		printf("%d\n",ans);
	}
	return 0;
}