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: Runtime Error)

#include<stdio.h>
int main()
{
int A[1000000000],N,n, i, sum=0;
scanf("%d",&N);
if(N>=2&&N<=2000)
{
for(i=1;i<=N;i++)
  {
  scanf("%d", &A[i]) ;
  }
  n=1;
for(i=1;i<=N;i++) 
{
if(n>=A[i])
  sum=sum+n*(A[i]-n) ;
  n++;
else
  sum=sum+n*(n-A[i]) ;
  n++;
}
printf("%d",sum);
}
}