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;

const int mod = 1e9 + 7;

long long rec(int left, int right, vector<vector<long long>>& dp, vector<pair<int, int>>& infant){
  if(left > right) return 0;
  if(dp[left][right] != -1) return dp[left][right];
  
  long long l = rec(left + 1, right, dp, infant);
  long long r = rec(left, right - 1, dp, infant);
  
  l += (long long)infant[right - left].first * abs(infant[right - left].second - left);
  r += (long long)infant[right - left].first * abs(infant[right - left].second - right);
  
  if(l < r){
    dp[left][right] = r;
  }else{
    dp[left][right] = l;
  }
  
  return dp[left][right];
}

int main(){
  int N;
  cin >> N;
  
  vector<vector<long long>> dp(N, vector<long long>(N, -1));
  vector<pair<int, int>> infant(N);
  long long joy = 0;
  for(int i = 0; i < N; i++){
    int A;
    cin >> A;
    infant[i] = make_pair(A, i);
  }
  
  sort(infant.begin(), infant.end());
  
  cout << rec(0, N - 1, dp, infant) << endl;
  
  return 0;
}