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: Time Limit Exceeded)

N = int(input())
A = list(map(int,input().split()))
B = [i for i in range(N)]
C = []
for i in range(N):
    C.append([A[i],B[i]])
C = sorted(C,reverse=True)
dp = [[0] * (N+1) for i in range(N+1)]
dp[0][0] = 0
for i in range(1,N+1):
    dp[i][0] = dp[i-1][0] + C[i-1][0]*(-C[i-1][1]+N-i)
    dp[0][i] = dp[0][i-1] + C[i-1][0]*(C[i-1][1]-i+1)
for i in range(1,N+1):
    for j in range(1,N+1):
        if i+j <= N:
            c = C[i+j-1][0]
            d = C[i+j-1][1]
            dp[i][j] = max(dp[i-1][j]+c*abs(N-i-d),dp[i][j-1]+c*abs(j-d-1))
        else:
            dp[i][j] = max(dp[i-1][j],dp[i][j-1])
print(dp[N][N])