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)

n = int(input())
a = list(map(int, input().split()))

a = list(enumerate(a))
a.sort(key = lambda x: x[1], reverse=True)

for i in range(n):
    a[i][0] = a[i][0] + 1

DP = [[0 for i in range(n+1)] for j in range(n+1)]
DP[0][0] = 0
for i in range(1, n+1):
    DP[0][i] = DP[0][i-1] + abs(a[i-1][1] * (n - i + 1 - a[i-1][0]))
    DP[i][0] = DP[i-1][0] + abs(a[i-1][1] * (a[i-1][0] - i))
    for x in range(1, i):
        y = i - x
        DP[x][y] = max(DP[x-1][y] + abs(a[i-1][1] * (a[i-1][0] - x)),  \
            DP[x][y-1] + abs(a[i-1][1] * (n - y + 1 - a[i-1][0])))

ans = 0
for i in range(n+1):
    ans = max(ans, DP[i][n-i])

print(ans)