CodeForces 703B

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:

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 every 1 ≤ 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 between a 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.

思路

一开始打算先设置城市数组，并且在录入的时候将城市之间的连接路段计算总数。然后再设置省会数组，每次录入将所有的城市遍历一次，发现时间复杂度太高不可行。

画了一会儿图之后发现，省会需要与除自己外所有城市连接，普通城市需要与所有省会连接。那我可以干脆先开两个sum1，sum2将所有城市加起来，所有省会加起来。（因为 [城市1 * 城市2 + 城市1 * 城市3 = 城市1 * (城市2 + 城市3) ] ）

然后录入省会是作标记，将所有城市遍历一次，判断是否省会，并且根据不同的类型算出他与其他城市相连的总和。然而这个得数应该是把我算进去的路都算了两次（1连2一次，2连1又一次），所以需要除以2。

最后再将相邻城市的路连接算上（两个城市都不是省会才需要计算，否则在前面已经算过了）。

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int city[100000+5]; //记录所有城市
int cap[100000+5];  //记录省会城市
int tr[100000+5];   //判断是否省会

int main(){
	int n,k,mark = 1;
	ll sum1 = 0,sum2 = 0,total = 0;
	cin >> n >> k;
	for(int i=1;i<=n;i++) {
		cin >> city[i];
		sum1 += city[i];
	}
	for(int i=1;i<=k;i++) {
		cin >> cap[i];
		tr[cap[i]] = 1;
		sum2 += city[cap[i]];
	}
	for(int i=1;i<=n;i++) {
		if(tr[i] == 1) {
			total += city[i] * (sum1 - city[i]);
		}
		else {
			total += city[i] * sum2;
		}
	}
	total /= 2;
	for(int i=1;i<n;i++) {
		if(tr[i] != 1 && tr[i+1] != 1) {
			total += city[i] * city[i+1];
		}
	}
	if(tr[1] != 1 && tr[n] != 1) {
		total += city[1] * city[n];
	}
	cout << total << endl;
}

题目来源

CodeForces 703B Mishka and trip