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)

import sys
stdin = sys.stdin
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
nn = lambda: list(stdin.readline().split())
ns = lambda: stdin.readline().rstrip()

n = ni()
a = na()
b = []
r = [0]
for i in range(n):
    b.append((a[i],i))
for i in a:
    r.append(r[-1]+i)

b.sort(reverse=True)

ans = 0

def f(left,right,k,c):
    if k == n:
        return c
    i,j = b[k]
    p = r[j]
    q = r[n]-r[j+1]
    x = (j-left)*i+p
    y = (right-j)*i+q
    if x >= y:
        return f(left+1,right,k+1,c+(j-left)*i)
    elif x < y:
        return f(left,right-1,k+1,c+(right-j)*i)

ans = f(0,n-1,0,0)

print(ans)