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>
using namespace std;
typedef  long long int ll;
#define pb push_back
typedef long double ld;
#define fi first
#define se second
#define mp make_pair

vector<vector<pair<ll,ll>>> out(1e5);
vector<vector<ll>> in(1e5);
ll inf=1e18;
ll p=1e9+7;
ll phi(ll n) {
    ll result = n;
    for (ll i = 2; i * i <= n; i++) {
        if(n % i == 0) {
            while(n % i == 0)
                n /= i;
            result -= result / i;
        }
    }
    if(n > 1)
        result -= result / n;
    return result;
}
ll power(ll x, ll y, ll p)
{
    ll res = 1;
    x = x % p;
   while (y > 0)
    {
        if (y & 1)
            res = (res*x) % p;
        y = y>>1;
        x = (x*x) % p;
    }
    return res;
}
ll modInverse(ll n, ll p)
{
    return power(n, p-2, p);
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
	ll n,i,j,t,a,b,m,k,x,y,ctr,f;
	ld r;
	cin>>n;
	ll arr[n];
	pair<ll,ll>cmax[n],temp;
	for(i=0;i<n;i++)
	{
	    cin>>arr[i];
	    t=std::max(i,n-1-i);
	    
	    temp=mp(t*arr[i],i);
	    //cout<<t*arr[i]<<"\n";
	    cmax[i]=temp;
	    
	    
	}
	ll ctr1=0,ctr2=n-1,ans=0;
	for(i=0;i<n;i++)
	{
	    sort(cmax,cmax+n);
	    if(cmax[n-1].fi!=-1)
	    ans+=cmax[n-1].fi;
	   
	    j=cmax[n-1].se;
	    t=std::max(abs(ctr1-j),abs(ctr2-j));
	    
	    if(t==abs(ctr1-j))
	    {
	    
	    ctr1++;
	    if(ctr1>ctr2)
	    {
	    ctr2--;
	    ctr1--;
	    }
	    }
	    else
	    {
	    ctr2--;
	    if(ctr1>ctr2)
	    {
	        ctr2++;
	        ctr1++;
	    }
	    }
	    arr[j]=-1;
	    for(j=0;j<n;j++)
	    cmax[j].fi=-1;
	    for(j=0;j<n;j++)
	    {
	        if(arr[j]!=-1)
	        {
	             t=std::max(abs(ctr1-j),abs(ctr2-j));
	             
	            
	               temp=mp(t*arr[j],j);
	               //if(ctr1==2&&ctr2==3)
	               //cout<<temp.fi;
	               cmax[j]=temp;
	             
	        }
	    }
	    
	}
	cout<<ans;
	return 0;
	
	
}