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;

const int N = 2020;

long long dp[N][N];
pair<int,int> a[N];

int main(){
    int n;
    scanf("%d", &n);
    for(int i = 1;i<=n;i++){
        int x;
        scanf("%d", &x);
        a[i] = pair<int,int>(-x,i);
    }
    sort(a+1,a+n+1);
    for(int i = 0;i<N;i++){
        for(int j = 0;j<N;j++){
            if(i == 0) dp[i][j] = 0;
            else dp[i][j] = -1;
        }
    }
    for(int i = 1;i<=n;i++){
        for(int j = 0;j<i;j++){
            int x = i-1-j, y = j;
            dp[x+1][y] = max(dp[x+1][y],dp[x][y]-(a[i].first)*abs(a[i].second-(x+1)));
            dp[x][y+1] = max(dp[x][y+1],dp[x][y]-(a[i].first)*abs(a[i].second-(n-y)));
        }
    }
    long long ans = -1;
    for(int i = 0;i<=n;i++){
        ans = max(ans,dp[n-i][i]);
    }
    printf("%lld\n", ans);
}