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 C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>

int main(){
 
  int n;
  int a[2000];
  int b[2000];
  int count;
  int sam;
  int max =0;
  
  scanf("%d",&n);
 
   for(int i=0;i<n;i++) {
    scanf("%d",&a[i]);
  	b[i] = i;
  }
  
  for(int i=0;i<n-1;i++){
	for(int j=0;j<n-1;j++){
      count = 0;
     for(int k=0;k<n;k++) count += a[k] *b[k];
      if(count > max) max=count;
      sam=b[j];
      b[j]=b[j+1];
      b[j+1] = sam;
    }
  }
  
   printf("%d",max);
  
  for(int k=0;k<n;k++) printf("\n%d",b[k]);
  
  return 0;
}