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 Julia (1.4.0) (judge verdict: Accepted)

function solve()
    n = parse(Int, readline())
    a = [parse(Int, x) for x in split(readline())]

    z = [(a[i], i) for i=1:n]
    sort!(z, rev=true)

    dp = zeros(Int, (n+1, n+1))
    for s=1:n, i=0:s
        li = i
        ri = s-i
        l = if li > 0
            dp[li, ri+1] + z[s][1] * abs(li - z[s][2])
        else
            0
        end
        r = if ri > 0
            dp[li+1, ri] + z[s][1] * abs((n-ri+1) - z[s][2])
        else
            0
        end
        dp[li+1, ri+1] = max(l, r)
    end
    maximum([dp[i+1, n-i+1] for i=0:n])
end

println(solve())