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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

n = int(input())
a = list(map(int, input().split()))
b = [(i, a[i]) for i in range(n)]
left, right = 0, n-1

ans = 0
while True:
    happiness = [max(abs(c[0]-left), abs(right-c[0])) * c[1] for c in b]
    # max/argmax
    tmp, argmax = 0, 0
    for i in range(len(b)):
        if happiness[i] > tmp:
            tmp, argmax = happiness[i], i
    if abs(b[argmax][0]-left) > abs(right-b[argmax][0]):
        left += 1
    else:
        right -= 1
    ans += tmp
    del b[argmax]
    if left == right:
        break

ans += max(abs(b[0][0]-left), abs(right-b[0][0])) * b[0][1]
print(ans)