Years have passed since Bash dropped out of university to become a Pokenom trainer. The adventure was full of difficulties and hardship, but Bash overcame all obstacles and became the best Pokenom trainer, like no one ever was!

Today Bash is celebrating his $13$th anniversary of dropping out of university and becoming a Pokenom trainer.

For the celebration party, Bash decided to prepare cakes for his $N$ Pokenoms. Bash’s Pokenoms are numbered from $1$ to $N$, inclusive. There are $N$ different cakes. The $i$-th Pokenom wants to eat the $i$-th cake.

The cakes are made using some ingredients. Each ingredient is uniquely marked by a prime number between $1$ and $N$, inclusive.

The recipe of the $X$-th cake contains $k$ grams of ingredient $p$ iff $p^{k}$ divides $X$, and $p^{k+1}$ does not. In other words, let the prime factorization of $X$ be $X = p_{1}^{k_{1}} \times p_{2}^{k_{2}} \times \cdots \times p_{m}^{k_{m}}$, the recipe of the $X$-th cake contains $k_{1}$ grams of ingredient $p_{1}$, $k_{2}$ grams of ingredient $p_{2}$, …, $k_{m}$ grams of ingredient $p_{m}$.

Bash goes to a supermarket to buy ingredients. There, Bash realizes that the ingredients are very expensive. If Bash buys $k$ grams of ingredient $q$, Bash’s happiness decreases by $k^{2} \times C_{q}$.

If the $i$-th Pokenom sees that Bash buys enough ingredient for the $i$-th cake, the Pokenom’s happiness increases by $V_{i}$.

Please help Bash buy ingredients so that the total happiness of Bash and his $N$ Pokenoms is maximized!

Note that the $i$-th Pokenom just needs to see that Bash has enough ingredients for the $i$-th cake. So even if the amount of ingredients Bash buys is enough to make either the $x$-th cake or the $y$-th cake, but not both, the total happiness still increases by $V_{x} + V_{y}$.

For example, consider $N = 100$ and Bash buys $2$ grams of ingredient $2$, $1$ gram of ingredient $3$ and $1$ gram of ingredient $5$: Bash’s happiness decreases by $4 \times C_{2} + 1 \times C_{3} + 1 \times C_{5}$. Bash has enough ingredients for cakes $1..6, 10, 12, 15, 20, 30,$ and $60$. So the happiness of the Pokenoms increases by\[ V_{1} + V_{2} + \cdots + V_{6} + V_{10} + V_{12} + V_{15} + V_{20} + V_{30} + V_{60}. \]

-----Input-----
 - The first line contains one integer $N$ $(1 \leq N \leq 10^{4})$.
 - The second line contains $N$ integers $V_{1}, V_{2}, \ldots V_{N}$$(0 \leq V_{i} \leq 10^{4})$.
 - The third line contains $N$ integers $C_{1}, C_{2}, \ldots , C_{N}$$(0 \leq C_{i} \leq 10^{4})$. It is guaranteed that $C_{i} = 0$ if $i$ is not a prime.

-----Output-----
Print a single integer $B$ — the maximum total happiness of Bash and his Pokenoms.

-----Examples-----
Sample Input 1:
10
1 2 3 40 5 6 7 8 9 10
0 2 3 0 5 0 7 0 0 0
Sample Output 1:
51

Sample Input 2:
1
2207
0
Sample Output 2:
2207

Source: kattis | Difficulty: MEDIUM


Solutions: