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<iostream>

using namespace std;

int main(){
  long long n, a[2009], sm=0;
  cin >> n;
  for(int i=0;i<n;i++){
    cin >> a[i];
  }
  int left=0, right=n-1;
  while(left<=right){
    int il, ir;
    long long mxl=0, mxr=0; 
    for(int i=0;i<n;i++){
      if(mxl<=abs(i-left)*a[i]){
        mxl = abs(i-left)*a[i];
        il = i;
      }
    } 
    for(int i=n-1;i>=0;i--){
      if(mxr<=abs(i-right)*a[i]){
        mxr = abs(i-right)*a[i];
        ir = i;
      }
    }
    if(mxl>mxr){
      sm += mxl;
      a[il] = 0;
      left++;
    }
    else{
      sm += mxr;
      a[ir] = 0;
      right--;
    }
  }
  cout << sm << endl;
}