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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		long ans = 0;
		int ans2 = 0;

		Scanner sc = new Scanner(System.in);
		int  n = sc.nextInt();
//		long  k = sc.nextLong();
//		longd = sc.nextInt();
//
//		boolean f = false;
		int[] a = new int[n+1];
		for (int i = 1; i < n+1; i++) {
			a[i] = sc.nextInt();
		}
		int[] b = new int[n+1];
		int[] c = new int[n+1];
		for (int i = 1; i <n+1; i++) {
			b[i] = sc.nextInt();
		}
		for (int i = 1; i <n; i++) {
			c[i] = sc.nextInt();
		}

		int chash = 0;

		for(int i : a) {
			ans2 += b[i];
			if(chash+1 == i) {
				ans2 += c[chash];
			}
			chash = i;
		}


		System.out.println(ans2);

		sc.close();
	}
}