CodeForces - 500C New Year Book Reading

题目描述：

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. 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.

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.

题目大意：

给你n本书，每本书都有重量。对于每天看的书，都要把之前的书先取出。可以改变书的摆放位置，问m天中，你取出书的重量最少是多少？

解题报告：

1：怎么改变书的顺序使得最少呢？当只需要观看一本书时，把这本书放最上面是最少的，这是毫无疑问的。

2：我们假设有两本书book1, book2，重量为w1, w2。每天观看毫无疑问就是要么先看book1或者book2。假设先看book1再看book2。

3：无非就两种摆放顺序（从上到下）：@1 book1  book2   @2 book2 book1

4：@1：w1              @2：w2+w1。从这里可以看出先观看的书要先摆在最上面。如果更多书都是差不多的。

5：对于计算，只需要把开始order序列构造出。然后每次从b数组中取出一个数x，从order数组的开始位置遍历到x相等结束。然后把这整段向后移动一个位置，第一本书的位置赋为x。

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 500+10;
ll w[N], b[N<<1], order[N], dic[N];
int main(){
    ll n, m, cnt = 0, ans = 0;
    scanf("%lld%lld", &n, &m);
    for(ll i=0; i<n; ++i){
        scanf("%lld", w+i);
    }
    for(ll i=0; i<m; ++i){
        scanf("%lld", b+i);
        if(!dic[b[i]])order[cnt++] = b[i], dic[b[i]]++;
    }
    for(ll i=0; i<m; ++i){
        ll x = b[i], y, j = 0;
        for(; j<cnt; ++j){
            if(x == order[j]){
                y = order[j];
                break;
            }
            ans += w[order[j]-1];
        }
        for(ll k=j; k>=1; --k){
            order[k] = order[k-1];
        }
        order[0] = y;
    }
    printf("%lld\n", ans);
    return 0;
}