In the first sample test:

In Peter's first test, there's only one cycle with 1 vertex. First player cannot make a move and loses.

In his second test, there's one cycle with 1 vertex and one with 2. No one can make a move on the cycle with 1 vertex. First player can replace the second cycle with two cycles of 1 vertex and second player can't make any move and loses.

In his third test, cycles have 1, 2 and 3 vertices. Like last test, no one can make a move on the first cycle. First player can replace the third cycle with one cycle with size 1 and one with size 2. Now cycles have 1, 1, 2, 2 vertices. Second player's only move is to replace a cycle of size 2 with 2 cycles of size 1. And cycles are 1, 1, 1, 1, 2. First player replaces the last cycle with 2 cycles with size 1 and wins.

In the second sample test:

Having cycles of size 1 is like not having them (because no one can make a move on them).

In Peter's third test: There a cycle of size 5 (others don't matter). First player has two options: replace it with cycles of sizes 1 and 4 or 2 and 3.

• If he replaces it with cycles of sizes 1 and 4: Only second cycle matters. Second player will replace it with 2 cycles of sizes 2. First player's only option to replace one of them with two cycles of size 1. Second player does the same thing with the other cycle. First player can't make any move and loses.
• If he replaces it with cycles of sizes 2 and 3: Second player will replace the cycle of size 3 with two of sizes 1 and 2. Now only cycles with more than one vertex are two cycles of size 2. As shown in previous case, with 2 cycles of size 2 second player wins.

So, either way first player loses.

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.

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

/*

下面是自己的代码，在第十组样例T了，虽然没想出来更好的优化方案，但是此次练习赛至少没看题解。下次要更加努力

*/

#include <stdio.h>

#include <iostream>

#include <string.h>

#include <algorithm>

#define N 100010

using namespace std;

long long n,k,a[N],b[N],ok[N];

int main()

{

    //freopen("in.txt","r",stdin);

    memset(ok,,sizeof ok);

    scanf("%lld%lld",&n,&k);

    for(int i=;i<=n;i++)

        scanf("%lld",&a[i]);

    for(int i=;i<=k;i++)

    {

        scanf("%lld",&b[i]);

        ok[b[i]]=;//标记主城

    }

    long long s=;

    for(int i=;i<=k;i++)

    {

        for(int j=;j<=n;j++)

        {

            if(b[i]==j) continue;

            if(ok[j])

                s+=a[j]*a[b[i]];

            else

                s+=a[j]*a[b[i]]*;

        }

    }

    for(int i=;i<n;i++)

    {

        if(ok[i]||ok[i+])

            continue;

        s+=a[i]*a[i+]*;

    }

    if(ok[]==&&ok[n]==)

        s+=a[]*a[n]*;

    printf("%lld\n",s/);

    return ;

}

#include <iostream>

#include <cstring>

#include <cstdlib>

#include <cstdio>

#include <cmath>

#include <algorithm>

#include <queue>

#include <stack>

#include <iomanip>

#include <map>

#define INF 0x3f3f3f3f

using namespace std;

typedef long long LL;

int dp[]={};

LL a[]={};

LL b[];

int main()

{

    int n,k;

    LL sum=,num=,num2=;

    scanf("%d %d",&n,&k);

    memset(dp,,sizeof(dp));

    for(int i=;i<n;i++)

    {

        scanf("%I64d",&a[i]);

        num+=a[i];

    }

    for(int j=;j<k;j++)

    {

        scanf("%I64d",&b[j]);

        num2+=a[b[j]-];

    }

    for(int x=;x<k;x++)

    {

        dp[b[x]-]=;

        sum+=(num-a[b[x]-])*a[b[x]-];

        sum-=(num2-a[b[x]-])*a[b[x]-];

        num2-=a[b[x]-];

    }

    for(int q=;q<n-;q++)

    {

        if(dp[q]!=&&dp[q+]!=)

        {

            sum+=a[q]*a[q+];

        }

    }

        if(!dp[n-]&&!dp[])

        sum+=a[]*a[n-];

    cout<<sum<<endl;

}