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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) for (int i = 0; i < n; ++i)
#define ALL(a) (a).begin(), (a).end()
#define rALL(a) (a).rbegin(), (a).rend()
#define SZ(x) ((int)(x).size())
using ll = long long;
using ull = unsigned long long;
using P = pair<int, int>;
using llP = pair<ll, ll>;
using DoP = pair<double, double>;
const int di[] = {0, 1, 0, -1, 1, 1, -1, -1};
const int dj[] = {1, 0, -1, 0, 1, -1, 1, -1};
const int INF = 1 << 28;
const ll INF64 = 1ll << 55;
const int mod = 1000000007;
//const int mod = 998244353;

template <class T>inline bool chmin(T &a, const T b)
{
  if (a > b)
  {
    a = b;
    return true;
  }
  return false;
}
template <class T>inline bool chmax(T &a, const T b)
{
  if (a < b)
  {
    a = b;
    return true;
  }
  return false;
}
template <class T>inline void line_out(const vector<T> vec) {
  int n = SZ(vec);
  rep(i, n) {
    cout << vec[i];
    if(i < n-1) cout << " ";
  }
  cout << endl;
  return;
}

struct mint
{
  ll x; // typedef long long ll;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const { return mint(-x); }
  mint &operator+=(const mint a)
  {
    if ((x += a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator-=(const mint a)
  {
    if ((x += mod - a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator*=(const mint a)
  {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const
  {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint a) const
  {
    mint res(*this);
    return res -= a;
  }
  mint &operator++()
  {
    if ((x += 1) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator--()
  {
    if ((x += mod - 1) >= mod)
      x -= mod;
    return *this;
  }
  mint operator*(const mint a) const
  {
    mint res(*this);
    return res *= a;
  }
  mint pow(ll t) const
  {
    if (!t)
      return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1)
      a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const
  {
    return pow(mod - 2);
  }
  mint &operator/=(const mint a)
  {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const
  {
    mint res(*this);
    return res /= a;
  }
};

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;
  rep(i, n) {
    ans += b[a[i]-1];
    if(i < n-1 && a[i+1]-a[i] == 1) ans += c[a[i]-1];
  }
  cout << ans << endl;
}