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)

# coding: utf-8
N=int(input())
A=list(map(int,input().split()))
INF=10**9
dp=[[-INF for i in range(N+1)] for j in range(N+1)]
dp[0][0]=0
AI=[]

for i in range(N):
    AI.append([A[i],i+1])
    
AI.sort()
AI.reverse()

for i in range(1,N+1):
    a=AI[i-1][0]
    I=AI[i-1][1]
    for j in range(N+1):
        if j==0:
            dp[i][j]=dp[i-1][j]+abs(I-(N+1-i))*a
        elif j==i:
            dp[i][j]=dp[i-1][j-1]+abs(I-j)*a
        else:
            dp[i][j]=max(dp[i-1][j-1]+abs(I-j)*a,dp[i-1][j]+abs(I-(N+1-(i-j)))*a)
ans=0
for i in range(N+1):
    ans=max(ans,dp[N][i])

print(ans)