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: Accepted)

#include <bits/stdc++.h>

using namespace std;
inline void prep ()
{
  cin.tie (0);
  cin.sync_with_stdio (0);
};
const double pi = acos(-1.);
//printf(%.12lf\n,mid);
// __builtin_popcount(int) count 1's in binary
// get decimal part of number
/*
double intpart;
double thiss = modf((double)(n*m)/(double)k, &intpart);
*/
// round = (int)(num+.5)
int mod = 998244353;

long long mymod(long long num){
  if (num< 0 && !(num%mod == 0)){
    if (mod == 1){
      return 0;
    }
    return mod-abs(num)%mod;
  }else{
    return num%mod;
  }
}

/*string sentence = And I feel fine...;
    istringstream iss(sentence);
*/  

double eps = .000000001;

int main ()
{
  prep();
  int n;
  cin >> n;
  int order[n];
  for (int i=0; i<n; i++){
      cin >> order[i];
  }
  int summ = 0;
  for (int i=0; i<n; i++){
      int num;
      cin >> num;
      summ += num;
  }
  int extra[n];
  for (int i=0; i<n-1; i++){
      cin >> extra[i];
  }
  for (int i=1; i<n; i++){
      if (order[i]-order[i-1] == 1){
          summ += extra[order[i-1]-1];
      }
  }
  cout << summ;
  return 0; 
}