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;

//g++ -g -o yourexe yourfile.cpp
//./yourexe < yourinput.in > youroutput.out

#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

#define M 1000000007

#define ll long long
//#define ld long double
#define vi vector<ll>
#define pi pair<ll, ll>
#define vii vector<pi>
#define vvi vector<vi>
#define pb push_back
#define endl "\n"
#define fi first
#define se second

#define REP(i, s, e) for (ll i = s; i < e; i++)
#define RREP(i, s, e) for (ll i = s; i > e; i--)

#define all(v) v.begin(), v.end()
#define part(v, s, e) v.begin() + s, v.begin() + e
#define print(v) for (auto i: v) cout << i << " ";

int main() {
    ll n;
    cin >> n;
    vi a(n), b(n), c(n-1);
    REP(i, 0, n)
        cin >> a[i];
    REP(i, 0, n)
        cin >> b[i];
    REP(i, 0, n-1)
        cin >> c[i];
    ll s = 0;
    s += b[a[0] - 1];
    REP(i, 1, n) {
        s += b[a[i] - 1];
        if (a[i] == a[i-1] + 1)
            s += c[a[i]-2];
    }
    cout << s << endl;
    return 0;
}