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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

N=int(input())
A=list(map(int,input().split()))

def f(A,N):
    H=list(range(N))
    S=0
    for i in range(N):
        if i%2:
            x=N-1-(i//2)
        else:
            x=i//2

        C=0
        L=[]
        M=-1
        for j in H:
            e=A[j]*abs(x-j)
            if M<e:
                M=e
                L=[j]
                C=1
            elif M==e:
                L.append(j)
                C+=1

        if i%2:
            u=L[(len(L)-1)//2]
        else:
            u=L[len(L)//2]
        H.remove(u)
        S+=M
    return S

B=[A[-k] for k in range(1,len(A)+1)]
print(max(f(A,N),f(B,N)))