codeforces B. Mishka and trip

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:

XXX consists of n cities, k of whose (just imagine!) are capital cities.
All of cities in the country are beautiful, but each is beautiful in its own way. Beauty value of i-th city equals to c_i.
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.
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.
There is at most one road between any two cities.
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 c_i·c_j.

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 c_a·c_b. 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, ..., c_n (1 ≤ c_i ≤ 10 000) — beauty values of the cities.

The third line of the input contains k distinct integers id₁, id₂, ..., id_k (1 ≤ id_i ≤ 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

input

5 2
3 5 2 2 4
1 4

output

Note

This image describes first sample case:

$\text{[math]}$

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

This image describes second sample case:

$\text{[math]}$

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

题意：有n个城市连成一个环，m个主要城市，每两个城市之间最多有一条路，每个主要城市都有通向其他所有城市的一条路，每条道路的权值是两端城市的值的乘积，求总的路线权值是多少。

分析：这种题都不会做了。。。比赛的时候竟然没有做出来。假设当前的主要城市是i，那么这个城市有通向其他所有城市的道路，算一下就是c[i]*(sum-c[i])，遍历所有的主要城市统计答案就行了，但是每一个主要城市的道路都被统计了两次，我们需要在统计的过程中不断把sum-c[i]，i是当前主要城市的id，最后主要把相邻的非主要城市的道路权值加上就行了。

#include<bitset>
#include<map>
#include<vector>
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<stack>
#include<queue>
#include<set>
#define inf 0x3f3f3f3f
#define mem(a,x) memset(a,x,sizeof(a))
#define F first
#define S second
using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

inline int in()
{
    int res=0;char c;int f=1;
    while((c=getchar())<'0' || c>'9')if(c=='-')f=-1;
    while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
    return res*f;
}
const int N=100010,MOD=1e9+7;
ll a[N];
bool vis[N];
int cap[N];


int main()
{
    int n=in(),m=in();
    ll sum=0;
    for(int i=0;i<n;i++){
        a[i] = in();
        sum += a[i];
    }
    for(int i=0;i<m;i++){
        cap[i]=in()-1;
        vis[cap[i]]=1;
    }
    ll ans=0;
    for(int i=0;i<n-1;i++){
        if(!vis[i] && !vis[i+1]) ans += a[i]*a[i+1];
    }
    if(!vis[0] && !vis[n-1]) ans += a[0]*a[n-1];
    for(int i=0;i<m;i++)
    {
        ans += (sum-a[cap[i]])*a[cap[i]];
        sum -= a[cap[i]];
    }
    cout<<ans<<endl;
    return 0;
}

codeforces B. Mishka and trip

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