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: Runtime Error)

from operator import itemgetter as get
N = int(input())

A = map(int, input().split())
SB = 0
ABI = sorted(((b:=max(N-i, i-1)*a, SB:=SB+b) and (a, b, i) for i, a in enumerate(A, 1)), reverse=True)

def solve(a, i, prev, th):
  pl, pr, ps = 0, -1, 0
  for l, r, s in prev:
    if s < th:
      continue
    if pr == r-1:
      yield l, r-1, max(s+abs(r-i)*a, ps+abs(i-pl)*a)
    else:
      if pl:
        yield pl+1, pr, ps+abs(i-pl)*a
      yield l, r-1, s+abs(r-i)*a
    pl, pr, ps = l, r, s
  yield pl+1, pr, ps+abs(i-pl)*a

prev = [(1, N, 0)]
pm = 0
for j, (a,b,i) in enumerate(ABI):
  M = 0
  prev = [(M:=max(M,s)) and (l, r, s) for l, r, s in solve(a,i, prev, pm-SB)]
  pm = M
  SB -= b

print(pm)