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<stdio.h>
#include<stdlib.h>
#define ll long long
int cmp(const void*a,const void*b){return*(long long*)a-*(long long*)b;}
int main(){
    ll n, i, j, ans = 0;
    scanf("%lld",&n);
    ll a[n + 1], b[n], c[n + 1], d[n + 1];
    for(i = 1; i <= n; i++)
        scanf("%lld",&a[i]);
    for(i = 1; i <= n; i++)
        b[i - 1] = a[i];
    for(i = 0; i <= n; i++)
        c[i] = 0;
    ll end = n, start = 1;
    qsort(b,n,sizeof(long long),cmp);
    for(i = 0; i < n; i++){
        for(j = 1; j <= n; j++){
            if(b[n - 1 - i] == a[j] && c[j] == 0){
                ll d1 = j - start;
                ll d2 = end - j;
                if(d1 < d2){
                    ans += a[j] * d2;
                    d[end] = a[j];
                    end --;
                }
                else{
                    ans += a[j] * d1;
                    d[start] = a[j];
                    start++;
                }
            }
        }
    }
    for(i = 1; i <= n; i++)
        printf("%lld ",d[i]);
    printf("\n");
    printf("%lld\n",ans);
    return 0;
}