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;

//C++14 (GCC 5.4.1)

#define LLINF (1LL<<60)
typedef long long ll;
#define mod 1000000007

#define repd(i,a,b) for (int i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
#define rrepd(i,a,b) for(int i=(a);i>=(b);i--)

template<typename T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<typename T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template < typename T > inline string toString( const T &a ) { ostringstream oss; oss << a; return oss.str(); }

#define dbg_v1(i,V) do{cout<<"-v1----\n";rep(i,V.size())cout<<" "<<i<<"\t:"<<V[i]<<endl;cout<<endl;}while(0)
#define dbg_v2(i,k,V) do{cout<<"-v2----\n";rep(i,V.size()){rep(k,V[0].size())cout<<" "<<V[i][k];cout<<endl;}}while(0)
#define lower_bound_idx(V,c) distance(V.begin(),lower_bound(V.begin(),V.end(),c))

int main() {
    ios::sync_with_stdio(false);//stdoutとcoutの同期解除
    cin.tie(nullptr);//cinとcoutの同期解除
    
    int N;
    cin >>N;

    vector<int> A(N),B(N),C(N-1);
    rep(i,N){
        cin >> A[i];
        A[i]--;
    }
    rep(i,N) cin >> B[i];
    rep(i,N-1) cin>>C[i];
    
    //-----------------------------------
    int ans=0;
    rep(i,N){
        ans+=B[A[i]];
            //cout<<"   "<<ans<<endl;
        if(i!=0){
            if(A[i-1]+1==A[i])ans+=C[A[i-1]];
        }
        //cout<<ans<<endl;
    }


    cout << ans <<endl;
    return (0);
}