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 int ll;
#define mod 1000000007
#include<string.h>
#define inf 9000000000000000000
#define maxn 200006
typedef pair<ll,ll> pll;
typedef pair<int,int> pint;
#define endl '\n'
#define mapint_iterator map<int,int> :: iterator
#define mapll_iterator map<ll,ll> :: iterator
#define setint_iterator set<int> :: iterator
#define setll_iterator set<ll> :: iterator
#define setpint_iterator set<pint> :: iterator
#define setpll_iterator set<pll> :: iterator
#define trace(x) cout<<#x<<": "<<x<<" "<<endl
#define trace2(x,y) cout<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define trace3(x,y,z) cout<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
#define trace4(a,b,c,d) cout<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl
#define cout1(a) cout<<a<<endl
#define cout2(a,b) cout<<a<<' '<<b<<endl
#define cout3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl
#define cout4(a,b,c,d) cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl
#define vcout(v,i) cout<<v[i].fi<<" "<<v[i].se<<endl
typedef priority_queue<ll,vector<ll>,greater<ll> > pqset;
typedef priority_queue<pint,vector<pint>,greater<pint> > pqset_ll;
#define pb push_back
#define fi first
#define se second
#include<ext/pb_ds/assoc_container.hpp>
ll dxk[]={0,0,1,-1,1,1,-1,-1};
ll dyk[]={1,-1,0,0,1,-1,1,-1};
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> os;
typedef tree<pll,null_type,less<pll>,rb_tree_tag,tree_order_statistics_node_update> os_pair;
#define acc (ios::sync_with_stdio(false),cin.tie(0))
#define rep(i,n) for(ll i=0;i<n;i++)
#define per(i,n) for(ll i=n-1;i>=0;i--)
#define rep1(i,n) for(ll i=1;i<=n;i++)
#define per1(i,n) for(ll i=n;i>0;i--)
#define repeat(i,start,n) for(ll i=start;i<n;i++)
#define print(a,n) for(ll i=0;i<n;i++){ cout<<a[i]<<" ";}
ll ll_max(ll a,ll b,ll c){return max(a,max(b,c));}
int int_max(int a,int b,int c){return max(a,max(b,c));}
ll ll_min(ll a,ll b,ll c){return min(a,min(b,c));}
int int_min(int a,int b,int c){return min(a,min(b,c));}
ll max(int a,ll b){ return max((ll)a,b);}
ll min(int a,ll b){ return min((ll)a,b);} 
ll min(ll a,int b){ return min(a,(ll)b);}
ll max(ll a,int b){ return max(a,(ll)b);}
ll dx[]={0,0,1,-1};
ll dy[]={1,-1,0,0};
//ifstream cin("b_read_on.txt"); ofstream cout("output3.txt");
//Chronology: Binary Search,Greedy,dp,dfs and similar
//Use (<<) for multiplication
//Use (>>) for division
// In dp, first try top-down,then bottom-up
//ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);cout<<fixed;cerr.tie(NULL);cout.precision(10);
// find_by_order -> value at index
// order_of_key -> index of value
ll n;vector<pll> v;
ll dp[2001][2001];
ll func(int i,int start,int end){
	if(i==n)
		return 0;
	if(dp[start][end]!=-1)
		return dp[start][end];
	ll o=func(i+1,start+1,end)+v[i].fi*abs(v[i].se-start);
	ll o1=func(i+1,start,end-1)+v[i].fi*abs(v[i].se-end);
	dp[start][end]=max(o,o1);
	return max(o,o1);
}
void solve(int countu){
	cin>>n;
	rep(i,n){
		ll b;
		cin>>b;
		v.pb({b,i});
	}
	sort(v.begin(),v.end());
	reverse(v.begin(),v.end());
	memset(dp,-1,sizeof(dp));
	cout<<func(0,0,n-1);
}
int main(){
    int t=1;
	//scanf("%d",&t);
	int countu=1;
    while(t--){
    	solve(countu);
    	countu++;
    }
}