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>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
int acs(const void *a, const void *b){return *(int*)a - *(int*)b;} /* 1,2,3,4.. */
int des(const void *a, const void *b){return *(int*)b - *(int*)a;} /* 8,7,6,5.. */
#define min(a,b) (a < b ? a: b)
#define max(a,b) (a > b ? a: b)

#define MAXN (20)
#define MOD (998244353)

int a[MAXN];
int b[MAXN];
int c[MAXN];

int main(void)
{
    int n;
    scanf("%d",&n);

    for(int i=0;i<n;i++) scanf("%d",&(a[i]));
    for(int i=0;i<n;i++) scanf("%d",&(b[i]));
    for(int i=0;i<n-1;i++) scanf("%d",&(c[i]));

    int ans = 0;
    for(int i=0;i<n;i++)
    {
        ans += b[a[i]-1];
        if(i>=1 && a[i]==(a[i-1]+1))
        {
            ans += c[a[i-1]-1];
        }
    }

    printf("%d\n",ans);
    return 0;
}