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: Accepted)

n = int(input())
a = list(map(int,input().split()))
ans = 0
place = [0]*n
active = sorted(a,reverse=True)
for i in range(n):
    place[i] = a.index(active[i])
    a[a.index(active[i])] = -1

#dp[x][y] x,y:0~n+1 (x[-1]=x[n+1]=0にすることで端の処理を楽にする)
dp = [[0 for i in range(n+2)]for j in range(n+2)]

for i in range(n):
    dp[0][i+1] = dp[0][i]+active[i]*abs(n-i-1-place[i])
    dp[i+1][0] = dp[i][0]+active[i]*abs(place[i]-i)



def check(mat):
    for i in range(len(mat)):
        print(mat[i])
    print("\n")


for i in range(n):
    for x in range(i+2):
        y = i+1 - x
        if x != 0 and y != 0:
            dp[x][y] = max(dp[x-1][y]+active[i]*abs(place[i]-x+1),dp[x][y-1]+active[i]*abs(n-y-place[i]))
            if i == n-1:
                ans = max(ans,dp[x][y])

print(ans)