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)

//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

#define Q int t; scanf("%d", &t); for(int q=1; q<=t; q++)
typedef long long int lli;
typedef pair<int, int> pi;
typedef pair<int, pi> node;
#define oset tree<pi, null_type,greater<pi>, rb_tree_tag,tree_order_statistics_node_update>


int main()
{
    int n;
    cin >> n;
    int a[n], b[n], c[n];
    for(auto &x: a)cin >> x, x--;
    for(auto &x: b) cin >> x;
    for(int i=0; i<n-1; i++){
        cin >> c[i];
    }
    int ans = 0;
    for(int i=0; i<n; i++){
        ans += b[a[i]];
        if(i>0 && a[i-1]+1==a[i]){
            ans += c[a[i-1]];
        }
    }
    cout << ans << "\n";

    return 0;
}