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;

#define rep(i, n) for (int i = 0; i < n; ++i)
#define stl_rep(itr, x) for (auto itr = x.begin(); itr != x.end(); ++itr)
#define stl_repr(itr, x) for (auto itr = x.rbegin(); itr != x.rend(); ++itr)
#define all(x) x.begin(), x.end()
#define allr(x) x.rbegin(), x.rend()

using llong = long long int;

const static int MOD = 1000000007;
const static int inf = INT_MAX;
const static llong INF = LLONG_MAX;
const static int dx[4] = {1, 0, -1, 0};
const static int dy[4] = {0, 1, 0, -1};

int main (int argc, char *argv[]) {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n;
    cin >> n;
    vector<int> A(n), B(n), C(n);
    rep(i, n) cin >> A[i];
    rep(i, n) cin >> B[i];
    rep(i, n-1) cin >> C[i];

    int satisfaction = 0;
    rep(i, n) {
        satisfaction += B[A[i]- 1];
        if (i < n - 1) {
            if (A[i + 1] == A[i] + 1) satisfaction += C[A[i] - 1];
        }
    }

    cout << satisfaction << endl;

    return 0;
}