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: Wrong Answer)

#include<bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < n; ++i)
#define rep2(i,s,n) for (int i = s; i < n; ++i)
#define all(a) a.begin(),a.end()
#define tmax(a,b,c) max(a,max(b,c))
#define tmin(a,b,c) min(a,min(b,c))
#define pb push_back
#define fi first
#define se second
using namespace std;
using ll = long long;
using P = pair<int,int>;
template<class T> inline bool chmin(T& a,T b) {if(a > b){a = b; return true;} return false;}
template<class T> inline bool chmax(T& a,T b) {if(a < b){a = b; return true;} return false;}
const int inf = 1001001001;

int main() {
    int n;
    cin >> n;
    vector<int> a(n),b(n),c(n-1);
    rep(i,n) cin >> a[i];
    rep(i,n) cin >> b[i];
    rep(i,n-1) cin >> c[i];
    int ans = 0;
    int now = -2;
    rep(i,n) {
        ans += b[a[i]];
        if(a[i] == now+1) ans += c[i-1];
        now = a[i];
    }
    cout << ans << endl;
}