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)

#include <iostream>
#include <vector>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <math.h>
using namespace std;

int main(){
    cin.tie(0);
   	ios::sync_with_stdio(false);

    int N;
    cin>>N;

    int A[N+2]={0};
    int B[N+2];
    int C[N+1];

    int cnt=0;

    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;i++){
        cin>>C[i];
    }


    for(int i=1;i<=N;i++){
        cnt=cnt+B[A[i]];

        if(A[i]+1==A[i+1]){
            cnt=cnt+C[A[i]];
        }


    }

    cout<<cnt<<"\n";

    return 0;

}