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 <iostream>
#include <vector>
#include <array>
#include <algorithm>
#include <cstring>
#include <queue>
#include <iomanip>
#include <numeric>
#include <cmath>

using namespace std;

//#define SIZE_OF_ARRAY(array) (sizeof(array)/sizeof(array[0]))

typedef long long ll;

//static const ll MOD = 1000000007;

int main() {
    ll N, answer = 0;

    // input
    cin >> N;
    vector<ll> A, B, C;
    for (ll i = 0; i < N; i++) {
        ll a;
        cin >> a;
        A.emplace_back(a - 1);
    }
    for (ll i = 0; i < N; i++) {
        ll b;
        cin >> b;
        B.emplace_back(b);
    }
    for (ll i = 0; i < N - 1; i++) {
        ll c;
        cin >> c;
        C.emplace_back(c);
    }

    // process
    for (ll i = 0; i < N; i++) {
        answer += B[A[i]];
        if (i >= 1 && A[i] == A[i - 1] + 1) {
            answer += C[A[i-1]];
        }
    }

    cout << answer << endl;

    return 0;
}