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_i
A = gets.split(" ").map(&:to_i)

arr = Array.new(N, nil)
N.times do |i|
  arr[i] = [i, A[i]]
end

arr.sort! { |a, b| b[1] <=> a[1] }
dp = Array.new(N+1) { Array.new(N+1, 0) }

N.times do |i|
  loc = arr[i][0]
  val = arr[i][1]
  0.upto(i) do |left|
    right = i - left

    dp[left+1][right] = [dp[left+1][right], dp[left][right] + ((loc - left) * val).abs].max
    dp[left][right+1] = [dp[left][right+1], dp[left][right] + ((loc - (N-right-1)) * val).abs].max
  end
end

res = 0
N.times do |left|
  right = N - left
  res = [res, dp[left][right]].max
end

puts res