C. New Year Book Reading

time limit per test  2 seconds

memory limit per test  256 megabytes

input  standard input

output  standard output

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 w_i.

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 b_j (1 ≤ b_j ≤ 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 aslifted 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, ..., w_n (1 ≤ w_i ≤ 100) — the weight of each book.

The third line contains m space separated integers b₁, b₂, ..., b_m (1 ≤ b_j ≤ 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 test(s)

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.

 

题意是n本书排成一堆，按给定的顺序读m次，每次读第bi本书。每本书有重量，读第bi本书的时候，代价就是所有在bi上面的书的总重量。看完后就直接放回书堆的最上层。要求确定一开始书的位置，使得代价尽量小

一开始我就是直接按每本书第一次出现的顺序搞的。比如样例1 3 2 3 1就直接确定为1 3 2  结果数据还真A了……结果最后因为小号交了一遍所以skipped……我真是无话可说……就是因为这个我从1900+掉到1800+了

后来想清楚了

假设我们考虑两本书i,j放的顺序对代价的影响，写一写就很容易发现影响只跟它第一次出现的顺序有关

比如一个看书的序列2 3 ……

有两种放法：2 3 ……或者3 2 ……

显然第一种的代价是w[2]，第二种的代价是w[2]+w[3]，显然第一种放法优

很容易发现按照2 3的顺序读书读完，它们的位置就唯一确定了。一定是3在2上面。后面也许还会出现要读2和3的情况，但是代价已经确定了

所以就是贪心完模拟啦

#include<cstdio>#include<iostream>#include<cstring>#include<cstdlib>#include<algorithm>#include<cmath>#include<queue>#include<deque>#include<set>#include<map>#include<ctime>#define LL long long#define inf 0x7ffffffusing namespace std;inline LL read(){    LL x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}int n,m,ans;int a[200010],b[200010];int lst[200010],nex[200010],succ[200010];int main(){    n=read();m=read();    for (int i=1;i<=n;i++)a[i]=read();    for (int i=1;i<=m;i++)b[i]=read();    for (int i=m;i>=1;i--)    {        nex[i]=lst[b[i]];        lst[b[i]]=i;    }    for (int i=1;i<=m;i++)    {        if (nex[i])succ[nex[i]]=i;    }    for (int i=1;i<=m;i++)    {        int x=succ[i]+1;        bool mrk[1010]={0};        for (int j=x;j<=i-1;j++)        {            if (!mrk[b[j]])ans+=a[b[j]],mrk[b[j]]=1;        }    }    printf("%d\n",ans);    return 0;}