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;
typedef long long ll;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(obj) (obj).begin(),(obj).end()
typedef priority_queue<int> pqmaxi;  //push(xx) top->maxi,pop();
const int mod=1e9+7;
typedef pair<ll,ll> P;
typedef tuple<int,int,int> TP;

template <class T>bool chmax(T& a, T b){
    if (a < b){
        a = b; 
        return 1; 
    } 
    return 0; 
    }

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;cin>>n;
  vector<P> a(n);
  ll tot=0;
  rep(i,n){
    cin>>a[i].first;
    a[i].second=i;
  }
  sort(a.begin(),a.end(),greater<P>());
  vector<vector<ll>> dp(n+1,vector<ll>(n+1,0));
  rep(i,n){
      for(int j=0;i+j<n;j++){
        chmax(dp[i+1][j],dp[i][j]+a[i+j].first*(a[i+j].second-i));
        chmax(dp[i][j+1],dp[i][j]+a[i+j].first*(n-1-j-a[i+j].second));
      }
  }

  ll res=0;
  rep(i,n+1)chmax(res,dp[i][n-i]);
  cout<<res<<endl;
  return 0;
  }