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: Runtime Error)

#include <iostream>
#include <cmath>
#include <cstdio>
#include <stdio.h>
#include <algorithm>
using namespace std;

const double pi = acos(-1);

typedef long long ll;
const int maxn = 2005;
struct pp{
    ll v;
    int id;
    bool operator<( const pp &o )const{
        if( v != o.v )return v > o.v;
        return id < o.id;
    }
};
pp A[maxn];
int mark[maxn];
ll dp[maxn][maxn];

int main()
{
    cin.tie(0);
    ios_base::sync_with_stdio(0);
    #ifdef acm
    freopen("a.in","r",stdin);
    #endif // acm

    int n;
    cin >> n;

    for( int i = 1; i <= n; i++ )
        cin >> A[i].v, A[i].id = i;

    sort(A+1, A+n+1);

    for( int i = 0; i <= n; i++ )
        for( int j = 0; j <= n; j++ )
            dp[i][j] = -1;

    dp[0][0] = 0;

    ll sol = 0;
    for( int i = 0; i <= n; i++ ){
        for( int j = 0; j <= n; j++ ){
            if( dp[i][j] == -1 )continue;
            int t = i+j+1;
            int ini = i+1;
            int fin = n-j;
            dp[i+1][j] = max( dp[i+1][j], dp[i][j] + abs(A[t].id - ini) * A[t].v );
            dp[i][j+1] = max( dp[i][j+1], dp[i][j] + abs(A[t].id - fin) * A[t].v );
        }
    }

    for( int i = 0; i <= n; i++ )
        sol = max(sol, dp[i][n-i]);

    cout << sol << "\n";

    return 0;
}