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>
int main() 
{
    long long int t, l, p, index, i, s, x, y, a[2100], sum=0,cnt=1,flag=0,flag1=0;
    scanf("%lli", &t) ;
    for(i=1;i<=t;i++) 
    scanf("%lli", &a[i]) ;
    
    
    s=0;
    for(l=1,p=t;cnt<=t;) 
   { for(i=1;i<=t;i++) 
    {
    	if(s<a[i]) 
    	{
    		s=a[i];
    		index=i;
    		if((index-l)>=(p-index)) 
    		{
    			x=index-l;
    			flag1=1;
    		} 
    		else
    		{
    			x=p-index;
    			flag1=-1;
    		} 
    	} 
    	else if(s==a[i]) 
    	{
    		if((i-l)>=(p-i)) 
    		{
    			y=i-l;
    			flag=1;
    		} 
    		else
    		{
    			y=t-i;
    			flag=-1;
    		} 
    		if(y>x) 
    		{
    			index=i;
    			x=y;
    			flag1=flag;
    		} 
    	} 
   } 
   sum+=(x*s);
   a[index]=0;
  if(flag1==1)
  l+=1;
  else if(flag1==-1)
  p-=1;
  
  cnt+=1;
  
   
   flag=flag1=0;
   s=0;
  } 
  printf("%lli\n", sum) ;
  return 0;
}