#P1519D. Maximum Sum of Products

    ID: 5898 Type: RemoteJudge 2000ms 256MiB Tried: 0 Accepted: 0 Difficulty: (None) Uploaded By: Tags>brute forcedpimplementationmathtwo pointers*1600

Maximum Sum of Products

No submission language available for this problem.

Description

You are given two integer arrays $a$ and $b$ of length $n$.

You can reverse at most one subarray (continuous subsegment) of the array $a$.

Your task is to reverse such a subarray that the sum $\sum\limits_{i=1}^n a_i \cdot b_i$ is maximized.

The first line contains one integer $n$ ($1 \le n \le 5000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($1 \le b_i \le 10^7$).

Print single integer — maximum possible sum after reversing at most one subarray (continuous subsegment) of $a$.

Input

The first line contains one integer $n$ ($1 \le n \le 5000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^7$).

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($1 \le b_i \le 10^7$).

Output

Print single integer — maximum possible sum after reversing at most one subarray (continuous subsegment) of $a$.

Samples

5
2 3 2 1 3
1 3 2 4 2
29
2
13 37
2 4
174
6
1 8 7 6 3 6
5 9 6 8 8 6
235

Note

In the first example, you can reverse the subarray $[4, 5]$. Then $a = [2, 3, 2, 3, 1]$ and $2 \cdot 1 + 3 \cdot 3 + 2 \cdot 2 + 3 \cdot 4 + 1 \cdot 2 = 29$.

In the second example, you don't need to use the reverse operation. $13 \cdot 2 + 37 \cdot 4 = 174$.

In the third example, you can reverse the subarray $[3, 5]$. Then $a = [1, 8, 3, 6, 7, 6]$ and $1 \cdot 5 + 8 \cdot 9 + 3 \cdot 6 + 6 \cdot 8 + 7 \cdot 8 + 6 \cdot 6 = 235$.