Codeforces 500C New Year Book Reading

New Year Book Reading

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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 ton. 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. 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.

Output

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

Sample Input

Input

3 5
1 2 3
1 3 2 3 1

Output

12

Hint

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

题目大意：有n各本具重量的书一本一本罗列在一起，m次阅读，每次抽取要阅读的书时需要搬动上面的所有书，求最终一共需要搬动的最少重量。
解题思路：书籍初始的罗列顺序是关键，只需按阅读顺序依次往下放置即使最优情况，然后模拟每次阅读时搬动的重量即可。
解题过程：用了类似与双向链表的方法模拟书籍在每个状态下的摆放序列。wa点：注意考虑多次看第一本书的情况。
代码如下：

#include <cstdio>
#include <cstring>
int w[10005];    //存放每本书的重量
int a[10005];    // 模拟的书籍摆放序列
int left[10005];
int right[10005];
int b[10010];   //存放编号，即阅读顺序
int vis[10005];
int main()
{
    int n,m,ans=0,i,x,xx,cnt=0,f;
    memset(vis,0,sizeof(vis));
memset(a,0,sizeof(a));
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)
        scanf("%d",&w[i]);
    for(i=0;i<m;i++)         // 输入编号即阅读顺序，构造初始摆放序列。
{
        scanf("%d",&b[i]);
if(!vis[b[i]])   
{
a[cnt++]=b[i];
vis[b[i]]++;
}
}
f=a[0]; // 模拟开始，把将要抽取的书籍上面书籍的重量累加，然后把书移动到最上面。
left[f]=0;
right[f]=a[1];
for(i=1;i<cnt;i++)
{
left[a[i]]=a[i-1];
right[a[i]]=a[i+1];
}
for(i=1;i<m;i++)
{
x=b[i];
xx=x;
while(true)
{
xx=left[xx];
if(!xx)
break;
ans+=w[xx];
}
if(x!=f)
{
right[left[x]]=right[x];
left[right[x]]=left[x];
left[x]=0;
right[x]=f;
left[f]=x;
f=x;
}
}
printf("%d\n",ans);
    return 0;
}