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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

"use strict";
var input=require("fs").readFileSync("/dev/stdin","utf8");
var cin=input.split(/ |\n/),cid=0;
function next(a){return a?cin[cid++]:+cin[cid++];}
function nexts(n,a){return a?cin.slice(cid,cid+=n):cin.slice(cid,cid+=n).map(a=>+a);}
function nextm(h,w,a){var r=[],i=0;if(a)for(;i<h;i++)r.push(cin.slice(cid,cid+=w));else for(;i<h;i++)r.push(cin.slice(cid,cid+=w).map(a=>+a));return r;}
function xArray(v){var a=arguments,l=a.length,r="Array(a["+--l+"]).fill().map(x=>{return "+v+";})";while(--l)r="Array(a["+l+"]).fill().map(x=>"+r+")";return eval(r);}

var myOut = main();
if(myOut !== undefined)console.log(myOut);

function main(){
  var n = next();
  var a = [];
  for(var i = 0; i < n; i++){
    a.push([next(),i]);
  }
  a.sort((a,b)=>b[0]-a[0]);
  var dp = xArray(0,n+1);
  for(var i = 0; i < n; i++){
    dp[i+1] = dp[i] + (n-i-a[i][1]-1) * a[i][0];
    for(var j = i; j; j--){
      dp[j] += (a[i][1]-(i-j)) * a[i][0];
      var t = dp[j-1] + (n-j-a[i][1]) * a[i][0];
      if(dp[j] < t)dp[j] = t;
    }
    dp[0] += (a[i][1]-i) * a[i][0];
  }
  return Math.max(...dp);
}