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;
#define int long long
#define ll long long
#define P pair<int,int>
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<n;i++)
#define all(v) v.begin(),v.end()
#define pb push_back
template<class T>void chmax(T &a,T b){if(a<b)a=b;}
template<class T>void chmin(T &a,T b){if(a>b)a=b;}
constexpr int INF=1000000000000000000;
constexpr int mod=1000000007;
int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int kaijo[2000010];
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a,b)*b;}
bool prime(int a){
    if(a==1)return false;
    for(int i=2;i*i<=a;i++){
        if(a%i==0)return false;
    }
    return true;
}
void init_fact(){
    kaijo[0]=1;
    for(int i=1;i<=2000000;i++){
        kaijo[i]=kaijo[i-1]*i;
        kaijo[i]%=mod;
    }
}
int modpow(int a,int b){
    if(b==0)return 1;
    if(b%2)return modpow(a,b-1)*a%mod;
    int memo=modpow(a,b/2);
    return memo*memo%mod;
}
int comb(int a,int b){
    if(!kaijo[0])init_fact();
    return kaijo[a]*modpow(kaijo[a-b],mod-2)%mod*modpow(kaijo[b],mod-2)%mod;
}
int inv(int x){return modpow(x,mod-2);}
bool kosa(double ax,double ay,double bx,double by,double cx,double cy,double dx,double dy){
    double ta=(cx-dx)*(ay-cy)+(cy-dy)*(cx-ax);
    double tb=(cx-dx)*(by-cy)+(cy-dy)*(cx-bx);
    double tc=(ax-bx)*(cy-ay)+(ay-by)*(ax-cx);
    double td=(ax-bx)*(dy-ay)+(ay-by)*(ax-dx);
    return tc*td<0&&ta*tb<0;
}
int n,dp[2010][2010];
P a[2010];
signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin>>n;
    rep(i,n){
        cin>>a[i].fi;
        a[i].se=i;
    }
    sort(a,a+n,greater<P>());
    rep(i,n){
        int p=a[i].se;
        for(int j=0;j<=i;j++){
            int le=j,ri=j+n-i-1;
            //dp[le][ri]
            chmax(dp[le+1][ri],dp[le][ri]+abs(p-le)*a[i].fi);
            chmax(dp[le][ri-1],dp[le][ri]+abs(p-ri)*a[i].fi);
        }
    }
    int ans=0;
    rep(i,n-1){
        chmax(ans,dp[i+1][i]);
    }
    cout<<ans<<endl;
    return 0;
}