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: Wrong Answer)

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <fstream>
#include <cassert>
#include <cstring>
#include <unordered_set>
#include <unordered_map>
#include <numeric>
#include <ctime>
#include <bitset>
#include <complex>
#include <chrono>
#include <random>
#include <functional>

using namespace std;

#define int long long

signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	int n;
	cin >> n;
	vector<int> a(n);
	for (auto &t : a) {
		cin >> t;
	}
	vector<int> ind(n);
	iota(ind.begin(), ind.end(), 0);
	while (true) {
		int fn = 0;
		for (int i = 0; i + 1 < n; i++) {
			auto f = abs(i - ind[i]) * a[ind[i]] + abs(i + 1 - ind[i + 1]) * a[ind[i + 1]];
			auto s = abs(i + 1 - ind[i]) * a[ind[i]] + abs(i - ind[i + 1]) * a[ind[i + 1]];
			if (f < s) {
				fn = 1;
				swap(ind[i], ind[i + 1]);
			}
		}
		if (!fn) break;
	}

	int ans = 0;
	for (int i = 0; i < n; i++) {
		ans += abs(i - ind[i]) * a[ind[i]];
	}
	cout << ans << endl;
}