cf #365b 巧妙的统计

 Mishka and trip

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Mishka is a great traveller and she visited many countries. After thinking about where to travel this time, she chose XXX — beautiful, but little-known northern country.

Here are some interesting facts about XXX:

1. XXX consists of n cities, k of whose (just imagine!) are capital cities.
2. All of cities in the country are beautiful, but each is beautiful in its own way. Beauty value of i-th city equals to ci.
3. All the cities are consecutively connected by the roads, including 1-st and n-th city, forming a cyclic route 1 — 2 — ... — n — 1. Formally, for every 1 ≤ i < n there is a road between i-th and i + 1-th city, and another one between 1-st and n-th city.
4. Each capital city is connected with each other city directly by the roads. Formally, if city x is a capital city, then for every1 ≤ i ≤ n,  i ≠ x, there is a road between cities x and i.
5. There is at most one road between any two cities.
6. Price of passing a road directly depends on beauty values of cities it connects. Thus if there is a road between cities i and j, price of passing it equals ci·cj.

Mishka started to gather her things for a trip, but didn't still decide which route to follow and thus she asked you to help her determine summary price of passing each of the roads in XXX. Formally, for every pair of cities a and b (a < b), such that there is a road betweena and b you are to find sum of products ca·cb. Will you help her?

Input

The first line of the input contains two integers n and k (3 ≤ n ≤ 100 000, 1 ≤ k ≤ n) — the number of cities in XXX and the number of capital cities among them.

The second line of the input contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 10 000) — beauty values of the cities.

The third line of the input contains k distinct integers id1, id2, ..., idk (1 ≤ idi ≤ n) — indices of capital cities. Indices are given in ascending order.

Output

Print the only integer — summary price of passing each of the roads in XXX.

Examples

input

4 1
2 3 1 2
3

output

17

input

5 2
3 5 2 2 4
1 4

output

71

Note

This image describes first sample case:

It is easy to see that summary price is equal to 17.

This image describes second sample case:

It is easy to see that summary price is equal to 71.

#include<iostream>
#include<stdio.h>
using namespace std;
const int maxx = ;
int city[maxx];
bool cap[maxx];
int cap_city[maxx];
long long val_total;
long long  total_cap;
int main()
{
    int n,k;
    val_total=;
    scanf("%d%d",&n,&k);
    for(int i=;i<=n;i++)
    {
        scanf("%d",city+i);
        val_total+=city[i];
    }
    total_cap=;
    for(int i=;i<k;i++)
    {
        int tmp;
        scanf("%d",&tmp);
        cap[tmp]=true;
        cap_city[i]=tmp;
        total_cap+=city[tmp];
    }
    city[]=city[n];
    cap[]=cap[n];
    long long ans=;
    for(int i=;i<=n;i++)
    {
        if(cap[i])
        {
            ans+=*(city[i]*1LL*(val_total-city[i])*1LL);
            if(!cap[(i-)%n])
            {
                ans-=city[i]*city[(i-)%n];
            }
            if(!cap[(i+)%n])
            {
                ans-=city[i]*city[(i+)%n];
            }
        }
        else
            ans+=(city[i]*city[i-]+city[i]*city[(i+)%n]);
    }
    for(int i=;i<k;i++)
    {
        ans-=(city[cap_city[i]]*(total_cap-city[cap_city[i]])*1LL);

    }
    printf("%I64d\n",ans/);
    return ;
}

简单统计所有边的代价是很简单的，但是因为点数较多，这样是会超时的。可以预先统计出所有点的总代价，所有关键点的总代价。计算时，将所有边统计两遍，其中非关键点较好处理，只要向两边连边，而关键点向所有点连两倍边，关键点之间会有四条边，要后期减去，而关键点和不是相邻的非关键点之间就是两倍边，而关键点和相邻的非关键点（如果是的话）是三倍边，要减去一次。最后，统计完之后，将总代价除以2，即为所求。