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: Wrong Answer)

import sys
readline = sys.stdin.readline

N = int(readline())
A = list(map(int, readline().rstrip().split()))
B = [-1] * N

def find_best():
    ma_sel = -1
    for i, a in enumerate(A):
        if a == -1:
            continue
        ma = -1
        for j in range(N):
            if B[j] != -1:
                continue
            get = a * abs(j-i)
            if get > ma:
                ma = get
                idx = j
        if ma > ma_sel:
            ma_sel = ma
            idx_from = i
            idx_to = idx

    return idx_from, idx_to, ma_sel

def main():
    global A, B
    for _ in range(N):
        idx_from, idx_to, ma_sel = find_best()
        A[idx_from] = -1
        B[idx_to] = ma_sel

    print(sum(B))


if __name__ == '__main__':
    main()