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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>

int main(void){

  typedef struct Ryouri{
    int num;
    int manzoku;
    int tuika;

  }ryouri;

  int n;
  int i,j;
  int ans=0;
  scanf("%d",&n);

  ryouri ry[n];
  int order[n];

  for(i=0;i<n;i++){
    ry[i].num=i;
  }

  for(i=0;i<n;i++){
    scanf("%d",&order[i]);
    order[i]=order[i]-1;
  }
  for(i=0;i<n;i++){
    scanf("%d",&ry[i].manzoku);
  }
  for(i=0;i<n-1;i++){
    scanf("%d",&ry[i].tuika);
  }

  /*for(i=0;i<n;i++){
    for(j=i+1;j<n;j++){
      if(ry[i].itu>ry[j].itu){
        tmp=ry[i];
        ry[i]=ry[j];
        ry[j]=tmp;
      }
    }
  }*/
  for(i=0;i<n;i++){
  ans=ans+ry[i].manzoku;
  }
  for(i=0;i<n-1;i++){

    if(ry[order[i]].num==ry[order[i+1]].num-1){
      ans=ans+ry[order[i]].tuika;
    }
  }
  printf("%d",ans);




return 0;

}