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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*

fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)
    val n = sc.nextInt()
    val a = (0 until 3).map {i ->
        (0 until n).map {j ->
            if (i == 2 && j == n - 1) {
                0
            } else {
                sc.next().toInt()
            }
        }.toMutableList()
    }
    println(problem140b(n, a))
}

fun problem140b(n: Int, a: List<MutableList<Int>>): Int {
    val dp = (0 until 2).map {
        (0..n).map {
            0
        }.toMutableList()
    }
    dp[1][0] = a[0][0]
    for (i in 1..n) {
        val tmp1 = a[1][a[0][i - 1] - 1]
        val tmp2 = dp[0][i - 1]
        dp[0][i] = tmp2 + tmp1
        dp[1][i] = a[0][i - 1]
        if (dp[1][i] == dp[1][i - 1] + 1) {
            dp[0][i] += a[2][dp[1][i - 1] - 1]
        }
    }
    return dp[0][n]
}