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: Wrong Answer)

n=int(input())
nums=list(map(int,input().split()))
ans=0
r=0
l=n-1
for i in range(n):
    maxh=0
    pos=0
    lr=0
    c=(l+r)/2
    for j in range(n):
        happy=nums[j]*max(j-r,l-j)
        if happy>maxh or (happy==maxh and nums[j]>nums[pos]):
            maxh=happy
            pos=j
    if c.is_integer() and pos==c:
        forsum=0
        for j in range(int(c)):
            forsum+=nums[j]
        backsum=sum(nums)-forsum-nums[int(c)]
        if forsum>=backsum:
            lr=1
        else:
            lr=2
    if pos>c or lr==1:
        r+=1
    else:
        l-=1
    ans+=maxh
    nums[pos]=0
print(ans)