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: Wrong Answer)


import sys
#import numpy as np

sys.setrecursionlimit(10 ** 9)
#def input():
#    return sys.stdin.readline()[:-1]

N = int(input())
#X, Y = map(int,input().split())
A = list(map(int,input().split()))

left = [0]
right = [0]
ans = 0
for i in range(N):
    l, r = 0, 0
    for j in range(len(left)):
        if A[i] > left[j]:
            buff = i - len(left) + 1 if left[j] == 0 else 1
            l += (A[i] - left[j]) * buff
        else:
            break
    for j in range(len(right)):
        if A[i] > right[j]:
            buff = N-i-1 - len(right) + 1 if right[j] == 0 else 1
            r += (A[i] - right[j]) * buff
        else:
            break
    if r > l:
        right.append(A[i])
        right.sort()
        ans += r
    else:
        left.append(A[i])
        left.sort()
        ans += l
#    print(i,A[i],l,r,ans)
#print(ans)
ans1 = ans

A.reverse()

left = [0]
right = [0]
ans = 0
for i in range(N):
    l, r = 0, 0
    for j in range(len(left)):
        if A[i] > left[j]:
            buff = i - len(left) + 1 if left[j] == 0 else 1
            l += (A[i] - left[j]) * buff
        else:
            break
    for j in range(len(right)):
        if A[i] > right[j]:
            buff = N-i-1 - len(right) + 1 if right[j] == 0 else 1
            r += (A[i] - right[j]) * buff
        else:
            break
    if r > l:
        right.append(A[i])
        right.sort()
        ans += r
    else:
        left.append(A[i])
        left.sort()
        ans += l
#    print(i,A[i],l,r,ans)
print(max(ans, ans1))