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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int[] A = new int[N];
    int[] B = new int[N];
    int[] C = new int[N-1];

    int ans = 0;
    for (int i = 0; i < N ; i++) {
      A[i]=sc.nextInt()-1;
    }
    for (int i = 0; i < N ; i++) {
      B[i]=sc.nextInt();
      ans += B[i];
    }
    for (int i = 0; i < N-1 ; i++) {
      C[i]=sc.nextInt();
    }

    for (int i = 1; i < N; i++) {
      int curMinus1 =A[i]-1;
      int prev = A[i-1];
      if( curMinus1 == prev ){
        ans += C[prev];
      }
    }


    System.out.println(ans);

  }



}