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;
typedef pair<int,int> pii;

const int N=2e3+5;
pii a[N];
ll dp[N][N];

void init();
void work();
int ca=1;
int main(){
//    freopen("in.txt","r",stdin);
    ios::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
//    init();
//    int T;cin>>T;
//    while(T--)
    work();
    return 0;
}
void work(){
    int n;cin>>n;
    int x;
    for(int i=1;i<=n;i++){
        cin>>x;
        a[i]=pii(x,i);
    }
    sort(a+1,a+n+1);
    for(int r=1;r<=n;r++){
        for(int l=r;l;l--){
            if(l==r) dp[l][r]=1LL*a[1].first*abs(l-a[1].second);
            else{
                int id=r-l+1;
                dp[l][r]=max(dp[l+1][r]+1LL*a[id].first*abs(a[id].second-l),
                             dp[l][r-1]+1LL*a[id].first*abs(a[id].second-r));
            }
        }
    }
    cout<<dp[1][n];
}