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: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <algorithm>
#include <iomanip>
#include <set>
#include <map>
#include <bitset>
#include <cmath>
#include <functional>

using namespace std;

#define REP(i,n) for(ll (i) = (0);(i) < (n);++i)
#define REV(i,n) for(ll (i) = (n) - 1;(i) >= 0;--i)
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define FI first
#define SE second
#define SHOW1d(v,n) {REP(WW,n)cerr << v[WW] << ' ';cerr << endl;}
#define SHOW2d(v,WW,HH) {REP(W_,WW){REP(H_,HH)cerr << v[W_][H_] << ' ';cerr << endl;}}
#define ALL(v) v.begin(),v.end()
#define Decimal fixed<<setprecision(20)
#define INF 1000000000
#define LLINF 1000000000000000000LL
#define MOD 998244353

typedef long long ll;
typedef pair<ll,ll> P;

ll dp[2222][2222];

int main(){
	cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);

	ll n;cin >> n;
	vector<P> a(n);
	REP(i, n) {
		cin >> a[i].FI;
		a[i].SE = i;
	}

	sort(ALL(a), greater<P>());

	REP(i, n+1){
		REP(j, n+1){
			if(i + j >= n)continue;
			ll left = i;
			ll right = n - 1 - j;
			ll id = a[i+j].SE;
			ll c = a[i+j].FI;
			dp[i+1][j] = max(dp[i+1][j], dp[i][j] + c * abs(id - left));
			dp[i][j+1] = max(dp[i][j+1], dp[i][j] + c * abs(id - right));
		}
	}

	ll ans = 0;
	REP(i, n)REP(j, n)if(i + j == n)ans = max(ans, dp[i][j]);
	cout << ans << endl;

	return 0;
}