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: Accepted)

#include<bits/stdc++.h>
using namespace std;
int main(){
  int N;
  cin>>N;
vector<pair<int64_t,int64_t>>A(N);
  for(int64_t i=0;i<N;i++){
  int64_t a; cin>>a;
    pair<int64_t,int64_t>p(a,i);
    A.at(i)=p;
  
  }sort(A.begin(),A.end());
  int64_t dp[N][N];
  for(int64_t j=0;j<N;j++)
    dp[0][j]=A[0].first*abs(j-A[0].second);
  for(int64_t i=1;i<N;i++){
  for(int64_t j=0;j<N-i;j++){
  dp[i][j]=
 max(dp[i-1][j]+A[i].first*abs(A[i].second-i-j),A[i].first*abs(A[i].second-j)+dp[i-1][j+1]);
  
  }
  
  }cout<<dp[N-1][0]<<endl;
   return 0;
}