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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,ans=0;
cin>>n;
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];
      }
    int d=a[n-1];
    ans+=b[d-1];
    for(int i=0;i<n-1;i++)
      {
        if(a[i]==n)
          {
            ans+=b[a[i]-1];

          }
          else
          {
            ans+=b[a[i]-1];
            ans+=c[a[i]-1];
          }

      }
      cout<<ans;

}