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>
using namespace std;
#define ALL(x) (x).begin(),(x).end()
#define COUT(x) cout<<(x)<<"\n"
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define REP(i, n) for(int i=0;i<n;i++)
#define YES(x) cout<<(x?"YES":"NO")<<"\n"
#define Yes(x) cout<<(x?"Yes":"No")<<"\n"
#define dump(x) cout<<#x<<" = "<<(x)<<"\n"
#define endl "\n"
using G = vector<vector<int>>;
using M = map<int,int>;
using P = pair<int,int>;
using PQ = priority_queue<int>;
using PQG = priority_queue<int,vector<int>,greater<int>>;
using V = vector<int>;
using ll = long long;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;

int main() {
  IOS;
  int n; cin >> n;
  int a[n],b[n],c[n];
  REP(i,n) { cin >> a[i]; a[i]--; }
  REP(i,n) cin >> b[i];
  REP(i,n-1) cin >> c[i];

  int ans = b[a[0]];
  for(int i = 1; i<n; i++) {
    ans += b[a[i]];
    if (a[i] - a[i-1] == 1) ans+=c[i-2];
  }
  COUT(ans);
  return 0;
}