CodeForces 703B Mishka and trip【水题】

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.

题意： 一个国家有n个城市，k个首都，城市是一个从1~n~1的一个环，首都通向各个城市。每两个城市间每条路的花费等于两个城市的val的城市，问你遍历所有的路的总花费。

#include <bits/stdc++.h>
#define MAXN 100005
using namespace std;
int main() {
    int n,m,k,cap[MAXN],c[MAXN] ;
    bool vis[MAXN];
    while (scanf("%d%d", &n, &k) != EOF) {
        __int64 sum = 0; memset(vis, false, sizeof(vis));
        //用sum记录所有城市的总的魅力值
        for (int i = 0; i < n; i++) {
            scanf("%d", &c[i]);
            sum += c[i];
        }
        __int64 ans = c[0]*c[n-1];
        //在没有首都的情况下
        for (int  i = 0; i < n - 1; i++) {
            ans += c[i]*c[i + 1];
        }
        for (int i = 0; i < k; i++) {
            scanf("%d", &m);
            if (m == 1) {
                //防止两个首都情况
                if (vis[n]) ans += c[m-1]*c[n-1];
                if (vis[m + 1]) ans += c[m]*c[m-1];
                ans += c[m-1]*(sum -c[m-1]-c[m]-c[n-1]);
            }
            else if (m == n) {
                //cout << ans << endl;
                if (vis[m-1]) ans += c[m-1]*c[m-2];
                if(vis[1]) ans += c[m-1]*c[0];
                //cout << ans << endl;
                ans += c[m-1]*(sum - c[m-1]-c[0]-c[m-2]);
            }
            else {
                if (vis[m-1]) ans += c[m-1]*c[m-2];
                if(vis[m+1]) ans += c[m-1]*c[m];
                ans += c[m-1]*(sum - c[m-1]-c[m-2]-c[m]);
            }
            vis[m] = true; sum -= c[m-1];
            //cout<<m<<" "<<sum<<" "<<ans<<endl;
        }
        printf("%lld
", ans);
    }
    return 0;
}