Problem:
Score : 200 points

Problem Statement

Takahashi went to an all-you-can-eat buffet with N kinds of dishes and ate all of them (Dish 1, Dish 2, \ldots, Dish N) once.

The i-th dish (1 \leq i \leq N) he ate was Dish A_i.

When he eats Dish i (1 \leq i \leq N), he gains B_i satisfaction points.

Additionally, when he eats Dish i+1 just after eating Dish i (1 \leq i \leq N - 1), he gains C_i more satisfaction points.

Find the sum of the satisfaction points he gained.

Constraints

All values in input are integers.

2 \leq N \leq 20

1 \leq A_i \leq N

A_1, A_2, ..., A_N are all different.

1 \leq B_i \leq 50

1 \leq C_i \leq 50

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N
B_1 B_2 ... B_N
C_1 C_2 ... C_{N-1}

Output

Print the sum of the satisfaction points Takahashi gained, as an integer.

Sample Input 1

3
3 1 2
2 5 4
3 6

Sample Output 1

14

Takahashi gained 14 satisfaction points in total, as follows:

First, he ate Dish 3 and gained 4 satisfaction points.

Next, he ate Dish 1 and gained 2 satisfaction points.

Lastly, he ate Dish 2 and gained 5 + 3 = 8 satisfaction points.

Sample Input 2

4
2 3 4 1
13 5 8 24
45 9 15

Sample Output 2

74

Sample Input 3

2
1 2
50 50
50

Sample Output 3

150

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*    Don’t Quit
    
    When things go wrong as they sometimes will;
        When the road you’re trudging seems all uphill;
            When the funds are low, and the debts are high;
                And you want to smile, but you have to sigh;
                    When care is pressing you down a bit
                        Rest if you must, but don’t you quit.
*/
#pragma gcc optimize ("unroll-loop")
#pragma gcc optimize ("Ofast")
#include <bits/stdc++.h>
using namespace std;
#define IOS ios_base::sync_with_stdio(false);cin.tie(0)
#define pb push_back
//#define mp make_pair
#define ff first
#define ss second
#define all(a) a.begin(),a.end()
#define int long long
const int MOD=1e9+7;
const int N=2e5+5;
int gcd(int n,int m)
{
    if(n==0)
        return m;
    else
        return gcd(m%n,n);
}
int a[N],b[N],c[N];
int32_t main()
{
    IOS;
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    for(int i=1;i<=n;i++)
    {
        cin>>b[i];
    }
    for(int i=1;i<=n-1;i++)
    {
        cin>>c[i];

    }
    int sum=0;
    for(int i=1;i<=n;i++)
    {
        if(a[i]==n)
        {
            sum+=b[a[i]];
        }
        else if(a[i]+1==a[i+1]&&i+1<=n)
        {   
            sum+=c[a[i]]+b[a[i]];
        }
        else
        {
            sum+=b[a[i]];
        }
    }
    cout<<sum<<endl;
    
         
}