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.
He lifts all the books above book x.
He pushes book x out of the stack.
He puts down the lifted books without changing their order.
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本书的重量，给你m天内这个人每天要看的书，他看书时，讲在要看的书上面的书搬走，看完后把书放在最上面，问你给一个书最初的排列，让他搬得总重量最小
分析
其实分析可得，我们最初放置的顺序是根据他的看书名单中所有书第一次出现的位置来决定的，是一个贪心，比如，他给你的序列是3111123331，那我们可以考虑，只有312放置可以让他最小。
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <cstdio>
using namespace std;
int b[510];
int x[1010];
vector<int>G;
bool vis[510];
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%d",&b[i]);
}
for(int i=0;i<m;i++)
{
scanf("%d",&x[i]);
if(!vis[x[i]])
{
vis[x[i]]=1;
G.push_back(x[i]);
}
}
int ans=0;
for(int i=0;i<m;i++)
{
for(int j=0;j<G.size();j++)
{
if(G[j]==x[i])
{
G.erase(G.begin()+j);
G.insert(G.begin(),x[i]);
break;
}
ans+=b[G[j]];
}
}
printf("%d\n",ans);
return 0;
}