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FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time. 

The treats are interesting for many reasons:

• The treats are numbered 1..N and stored sequentially in single file in a long box that is open at both ends. On any day, FJ can retrieve one treat from either end of his stash of treats.
• Like fine wines and delicious cheeses, the treats improve with age and command greater prices.
• The treats are not uniform: some are better and have higher intrinsic value. Treat i has value v(i) (1 <= v(i) <= 1000).
• Cows pay more for treats that have aged longer: a cow will pay v(i)*a for a treat of age a.

Given the values v(i) of each of the treats lined up in order of the index i in their box, what is the greatest value FJ can receive for them if he orders their sale optimally? 

The first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.

Input

Line 1: A single integer, N 

Lines 2..N+1: Line i+1 contains the value of treat v(i)

Output

Line 1: The maximum revenue FJ can achieve by selling the treats

Sample Input

5
1
3
1
5
2

Sample Output

43

Hint

Explanation of the sample: 

Five treats. On the first day FJ can sell either treat #1 (value 1) or treat #5 (value 2). 

FJ sells the treats (values 1, 3, 1, 5, 2) in the following order of indices: 1, 5, 2, 3, 4, making 1x1 + 2x2 + 3x3 + 4x1 + 5x5 = 43.

 

 

注意从小区间推大区间，从内向外推，也就是从后卖出的物品向前卖出的物品状态递推。

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

#define MAXN 2002
/*
区间DP,从后向前推，dp[i][j]表示首端元素为a[i],尾端为a[j]的情况
    dp[i][j] = max(dp[i+1][j]+t*a[i],dp[i][j-1]+t*a[j])
*/
int a[MAXN],dp[MAXN][MAXN];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
        dp[i][i] = n*a[i];//最后一个卖出元素是a[i]的情况
    for(int l=1;l<n;l++)
    {
        for(int i=1;i+l<=n;i++)
        {
            int j = i+l;
            dp[i][j] = max(dp[i+1][j]+(n-l)*a[i],dp[i][j-1]+(n-l)*a[j]);
        }
    }
    printf("%d
",dp[1][n]);
    return 0;
}