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 5.4.1) (judge verdict: Accepted)

#include "stdio.h"

int main()
{
	int N = 0;
	int A[21];
	int B[21];
	int C[21];
	int cnt;
	int sum = 0;


	
	scanf("%d", &N);
	
	
	for (cnt = 0; cnt < N; cnt++) {
		scanf("%d", &A[cnt]);

	}
	
	
	for (cnt = 0; cnt < N; cnt++) {
		scanf("%d", &B[cnt]);
	}
	
	for (cnt = 0; cnt < N - 1; cnt++) {
		scanf("%d", &C[cnt]);
	}



	for (cnt = 1; cnt <= N; cnt++) {
		sum = sum + B[A[cnt - 1] - 1];
		if (cnt - 1 >= 1) {
			if (A[cnt - 2] == A[cnt - 1] - 1) {
				sum = sum + C[A[cnt - 1] - 2];
			}

		}
		
	}
	printf("%d", sum);
	return 0;

}