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 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
#define Q int tt,qq; cin>>tt ;for(qq=1;qq<=tt;qq++)
#define FAST ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

int main()
{
    int n;
    cin>>n;
    vector<int>a(n),b(n),c(n-1);
    for(int i=0;i<n;i++)    cin>>a[i];
    for(int i=0;i<n;i++)    cin>>b[i];
    for(int i=0;i<n-1;i++)  cin>>c[i];
    for(int i=0;i<n;i++){
        a[i]=a[i]-1;
    }
    ll sum=0;
    for(int i=0;i<n;i++){
        sum+=b[a[i]];
        if(a[i+1]-a[i]==1)
            sum+=c[a[i]];
    }
    cout<<sum;

}