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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
int a[1000],b[1000],c[1000] = {0};
int cmp(const void *a,const void *b)
{
	return *(int *)a-*(int *)b;
}
int main(void)
{
	int n;
	int index = 0;
	scanf("%d",&n);
	int i;
	int count = 0;
	for(i = 1;i<=n;i++) scanf("%d",&a[i]);
	for(i = 1;i<=n;i++) {
		scanf("%d",&b[i]);
		count += b[i];
		}
	for(i = 1;i<n;i++) scanf("%d",&c[i]);
	qsort(c+1,n-1,sizeof(c[1]),cmp);
	for(i = 1;i<=n;i++)
	{
		if(a[i] < a[i+1] && index != n-1)
		{
			index++;
		}
	}
	for(i = 1;i<=index;i++)
		count += c[i];
	printf("%d",count);
	return 0;
}