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)

import numpy as np

N = int(input())
A_list = map(int, input().split())

# [活発度, もとの番号] のリストを降順に並べる
A_list = [[A, index] for index, A in enumerate(A_list)]
A_list.sort(reverse=True)

# print(A_list)

dp = np.zeros((N+1, N+1), int)
# print(dp)
for left in range(0, N):
    for right in range(0, N - left):
        # print(f'left, righte: { left, right }')
        # print(A_list[left + right - 1][0])
        activity = A_list[left + right][0]
        distance_from_left = A_list[left + right][1] - left
        # print(f'distance_from_left: { distance_from_left }')
        distance_from_right = (N - 1 - right) - A_list[left + right][1]
        # print(f'distance_from_right: { distance_from_right }')
        dp[left + 1][right]\
            = max(dp[left + 1][right],
                  dp[left][right] + activity * distance_from_left)
        dp[left][right + 1]\
            = max(dp[left][right + 1],
                  dp[left][right] + activity * distance_from_right)
        # print('dp---')
        # print(dp)
        # print('---dp')
# print(dp)
print(np.max(dp))