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 ll      long long int   
#define ld      long double
#define pii     pair<int ,int>
#define pld     pair<ld ,ld>
#define all(v)  v.begin(),v.end()
#define F       first
#define S       second
#define mod     1000000007
#define pb      push_back
#define MP      make_pair
#define mset(x) memset(x, 0, sizeof(x));
#define ios     ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define inf     1000000000000000007
const int N=300006;
ll n,i,ans,a[N],b[N],c[N];
void run(){
	cin>>n;
    for(i=1;i<=n;i++)
        cin>>a[i];
    for(i=1;i<=n;i++)
        cin>>b[i];
    for(i=1;i<=n-1;i++)
        cin>>c[i];
    for(i=1;i<=n;i++)
    {
        ans+=b[a[i]];
        if(i>=2 && a[i]==a[i-1]+1) ans+=c[a[i-1]]; 
    }
    cout<<ans;
 
}
 
 
void setup(){
    ios;
    //cout << setprecision(20);
    // #ifndef ONLINE_JUDGE
    // freopen("input", "r", stdin);
    // freopen("output", "w", stdout);
    // #endif
}
 
signed main(){
    setup();
    int tests = 1;
    //cin >> tests;
    for(int i = 1; i <= tests; i++) run();
    return 0;
}