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++ (Clang 10.0.0) (judge verdict: Accepted)

#include<iostream>
#include<vector>
#include<algorithm>
#include<functional>

using namespace std;

typedef long long ll;

int main() {
	ll n,tmp;
	ll ans=0;
	cin >> n;
	vector<pair<ll, ll>> children;
	int front = 1, end = n;
	for (int i = 1;i <= n;i++) {
		cin >> tmp;
		children.push_back(make_pair(tmp, i));
	}
	sort(children.begin(), children.end(),greater<pair<ll,ll>>());
	ll **dp = new ll*[n];
	for (int i = 0;i < n;i++) {
		dp[i] = new ll[i + 2];
	}
	//dp[0]の構成
	dp[0][0] = children[0].first * (children[0].second-1);
	dp[0][1] = children[0].first * (n - children[0].second);
	for (int i = 1;i < n;i++) {
		dp[i][0] = dp[i - 1][0] + children[i].first * abs(children[i].second - i - 1);
		for (int j = 1;j <= i;j++) {
			dp[i][j] = max(dp[i - 1][j] + children[i].first * abs(children[i].second - (i - j) - 1), dp[i - 1][j - 1] + children[i].first * abs(n - j  + 1 - children[i].second));
		}
		dp[i][i + 1] = dp[i - 1][i] + children[i].first * abs(n - i - children[i].second);
	}

	for (int i = 0;i <= n;i++) {
		ans = max(ans, dp[n - 1][i]);
	}
	cout << ans << endl;
}