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 <stdlib.h>
#include <string.h>
#include <math.h>
int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
//    return *(long long*)a - *(long long*)b;
}
//  qsort(A, b, sizeof(int), compare_int);

long long  factorial(long long n) {
    if (n > 0) {
        return n * factorial(n - 1);
    } else {
        return 1;
    }
}

int min(int a, int b) {
    if (a > b)
        return b;
    else
        return a;
}

int distance(int a, int b) {
  return abs(a-b)*abs(a-b);
}

//int ()



int main()
{
  int a ,b, c, d, e, X, Y;
//  long long  a, b, c, d, e, X, Y;
  //  double  c;
//  int  A[200003]={};
  int  A[100003]={};
  int  B[100003]={};
  int  C[100003][11]={};

  int  flg=0;

//  long long  a, b, c, d, X, Y;
  char s[101][101]={};
  char t[101]={};
//  int N, M;
//  int A[1000000001]={};
//  long long  C[101]={};
  int i,j,k;
  long long p,q;
//  int flag=0;
//  int tmp=0;
//  char tmp[101]={};
//  long long min=10000000000;
//  long long min=100000001;
  int len=0;
  int l=0;
  int r=1000001;
  int cnt=0;
  int num=0;
  int num1=0;
  int num2=0;
  int ans=0;
  int max=0;
  int sum=0;
  int addr=0;
  int ptn=0;
  long long  tmp=0;
//  bool on_off[100]={};
  
//  char alp[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

  //  a=0;
//  scanf("%lld %lld", &a, &b);
  scanf("%d", &a);

  scanf("%d", &A[0]);
  for(i=1;i<a;i++){
  scanf("%d", &A[i]);
    if(A[i-1]==A[i]-1)
    { B[A[i-1]-1]=1;
//  printf("%d", B[i]);
     
    }

  }
  for(i=0;i<a;i++){
    scanf("%d", &A[i]);
    cnt=cnt+A[i];
//    printf("%d ", cnt);
  }
  for(i=0;i<a-1;i++){
    scanf("%d", &A[i]);
    if(B[i]==1) cnt=cnt+A[i];
//    printf("%d ", cnt);
  }
 
  

    
//    printf("%s %d %d\n", s[i],C[i][0],C[i][1]);
   // printf("%d\n", C[i][1]);
/*  for(i=0;i<addr;i++){
//  scanf("%lld%lld", &B[0][i],&B[1][i]);
//    ans=ans+A[i];
    printf("%lld ", A[i]);
  }
*/
    printf("%d ", cnt);
   

return 0;
}