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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#define MIN(x,y) (((x) > (y)) ? (y) : (x))
#define MAX(x,y) (((x) < (y)) ? (y) : (x))
#define ABS(x) ((x) >= 0 ? (x) : (-(x)))

int a[2001];
long long dp[2001][2001];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      dp[i+1][j+1] = MAX(MAX(dp[i][j], dp[i][j+1]), dp[i+1][j]) + a[i] * ABS(i - j);
    }
  }
  printf("%lld", dp[n][n]);
  return 0;
}