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;
using ll = long long;
using db = double;
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define all(v) (v).begin(), (v).end()
#define siz(v) (ll)(v).size()
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define repi(i, x, n) for (ll i = x; i < (ll)(n); i++)
#define lb lower_bound
#define ub upper_bound
typedef pair<int, int> P;
typedef pair<ll, ll> PL;
const ll mod = 1000000007;
const ll INF = 1000000099;
vector<ll> dx = {-1, 1, 0, 0}, dy = {0, 0, -1, 1};
//cin.tie(0);ios::sync_with_stdio(false);


signed main()
{
    int n;cin>>n;
    ll ans=0;
    vector<ll> 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];
    }

    rep(i,n){
        ans+=b.at(a.at(i));
        if(i>0 && a.at(i-1)+1==a.at(i)){
            ans+=c.at(a.at(i-1));
        }
    }

    cout<<ans<<endl;
}