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<bits/stdc++.h>
#define ll long long 
using namespace std;
int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    ll n;
    cin>>n;
    vector<pair<ll,ll>> a(n);
    vector<ll> aa;
    for(int i=0;i<n;++i){
        cin>>a[i].first;
        a[i].second = i+1;
        aa.push_back(a[i].first);
    }
    ll vis[n+1]={0};
    ll mn = 1;
    ll mx = n;
    ll ans = 0;
    sort(a.rbegin(),a.rend());
    for(int i=0;i<n;++i){
        if(!vis[a[i].second]){
            if(abs(a[i].second-mn) >= abs(a[i].second-mx)){
                ans += a[i].first*abs(a[i].second-mn);
                mn = min(mn+1,n);
            }
            else{
                ans += a[i].first*abs(a[i].second-mx);
                mx = max(mx-1,0LL);   
            }
            ++vis[a[i].second];
        }
    }
    cout<<ans<<'\n';
    return 0;
}