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 <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

#define MOD 1000000007LL
const long long INF = 1LL << 60;
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))

typedef long long ll;

typedef struct Child{
    ll a;
    ll idx;
}child;

int ascend(const void *n,const void *m){
    if(((child *)n)->a > ((child *)m)->a)return -1;
    if(((child *)n)->a < ((child *)m)->a)return 1;
    if(((child *)n)->idx > ((child *)m)->idx)return -1;
    if(((child *)n)->idx < ((child *)m)->idx)return 1;
    return 0;
}
int N;
ll DP[2048][2048];
child A[2048];
ll assign(ll start,ll finish){
    if(start>finish) return 0;
    if(DP[start][finish]!=-1) return DP[start][finish];
    ll i = (start-1)+(N-finish);
    DP[start][finish] = MAX(assign(start+1,finish)+A[i].a*llabs(start-A[i].idx),DP[start][finish]);
    DP[start][finish] = MAX(assign(start,finish-1)+A[i].a*llabs(finish-A[i].idx),DP[start][finish]);
    return DP[start][finish];
}

int main(){
    scanf("%d",&N);
    for(int i=0;i<2048;i++)for(int j=0;j<2048;j++)DP[i][j]=-1;
    for(int i = 0;i<N;i++){
        scanf("%lld",&A[i].a);
        A[i].idx=i+1;
    }
    qsort(A,N,sizeof(A[0]),ascend);
    printf("%lld\n",assign(1,N));
    return 0;
}