Problem:
Score : 500 points

Problem Statement

There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.

You can rearrange these children just one time in any order you like.

When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.

Find the maximum total happiness points the children can earn.

Constraints

2 \leq N \leq 2000

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum total happiness points the children can earn.

Sample Input 1

4
1 3 4 2

Sample Output 1

20

If we move the 1-st child from the left to the 3-rd position from the left, the 2-nd child to the 4-th position, the 3-rd child to the 1-st position, and the 4-th child to the 2-nd position, the children earns 1 \times |1-3|+3 \times |2-4|+4 \times |3-1|+2 \times |4-2|=20 happiness points in total.

Sample Input 2

6
5 5 6 1 1 1

Sample Output 2

58

Sample Input 3

6
8 6 9 1 2 1

Sample Output 3

85

Solution in Python (3.8.2) (judge verdict: Accepted)


def main():
    n = int(input())
    a = list(map(int, input().split()))
    a = sorted([(j, i) for i, j in enumerate(a)])[::-1]
    dp = [0]
    for i in range(n):
        dp2 = [0]*(i+2)
        A, p = a[i]
        for j in range(i+1):
            right = n-i-1+j
            dp2[j] = max(dp2[j], dp[j]+abs(p-right)*A)
            dp2[j+1] = max(dp2[j+1], dp[j]+abs(p-j)*A)
        dp = dp2
    print(max(dp2))


main()