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 Ruby (2.7.1) (judge verdict: Accepted)

n = gets.to_s.to_i
a = gets.to_s.split.map{|t| t.to_i }

a = a.map.with_index{|t,i| [t,i] }.sort.reverse

dp = Array.new(n+1){ [0] * (n+1) }

a.each_with_index do |tmp,i|
  a = tmp[0]
  k = tmp[1]
  (i+1).times do |j|
    
    dp[i+1][j+1] = [ dp[i+1][j+1],  dp[i][j] + (k-j) * a       ].max
    dp[i+1][j]   = [ dp[i+1][j],    dp[i][j] + (n-i+j-1-k) * a ].max
  end
end

puts dp[-1].max