#include <bits/stdc++.h>

// definitions
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define rep(i, l, r) for (long long i = (l); i < (long long)(r); i++)
#define repb(i, r, l) for (long long i = (r); i > (long long)(l); i--)
#define pb push_back
#define mod1 998244353
#define mod 1000000007
#define MAX_CHAR 256

using namespace std;
typedef vector<long long> vl;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector< pair<long long, long long> > vp;
typedef vector<tuple<long long , long long , long long> > vtup;
typedef deque <long long> dq;
typedef long long int lli;
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;


// standard functions
ll gcd(ll x, ll y) {
    if (x < y) return gcd(y, x);
    if (y == 0) return x;
    return gcd(y, x % y);
} 
 


ll modexp(ll x, ll ex, ll modp) {
    ll ans = 1ll;
    while (ex > 0) {
        if (ex & 1ll) ans = (ans * x)%modp;
        ex >>= 1ll;
        x = (x * x)%modp;
    }
    return ans;
} 
 
ll nCrModp(ll n, ll r, ll p) 
{ 
    ll C[r+1]; 
    memset(C, 0, sizeof(C)); 
  
    C[0] = 1; 
    for (ll i = 1; i <= n; i++) 
    { 
        for (ll j = min(i, r); j > 0; j--) 
            C[j] = (C[j] + C[j-1])%p; 
    } 
    return C[r]; 
}
 
bool isPrime(ll n) 
{ 
    if (n <= 1)  return false; 
    if (n <= 3)  return true;  
    if (n%2 == 0 || n%3 == 0) return false; 
  
    for (ll i=5; i*i<=n; i=i+6) 
        if (n%i == 0 || n%(i+2) == 0) 
           return false; 
  
    return true; 
} 
 
vl countFreq(ll arr[], ll n) 
{ 
    unordered_map<ll, ll> mp; 
    vector <ll> v;
    
    for (ll i = 0; i < n; i++) 
        mp[arr[i]]++; 
   
    for (auto x : mp) 
        v.pb(x.second);
    
    return v;
}

ll binarySearch(ll arr[], ll l, ll r, ll x) 
{ 
    while (l <= r) 
    { 
        ll m = l + (r - l) / 2; 
        if (arr[m] == x) 
            return m; 
        if (arr[m] < x) 
            l = m + 1; 
        else
            r = m - 1; 
    }
    return -1; 
}

const ll MAXSIEVE=(1e3);
ll spf[MAXSIEVE];
void sieve() 
{ 
    spf[1] = 1; 
    for (ll i=2; i<MAXSIEVE; i++) 
        spf[i] = i; 

    for (ll i=4; i<MAXSIEVE; i+=2) 
        spf[i] = 2; 
  
    for (ll i=3; i*i<MAXSIEVE; i++) 
    { 
        if (spf[i] == i) 
        { 
            for (ll j=i*i; j<MAXSIEVE; j+=i) 
                if (spf[j]==j) 
                    spf[j] = i; 
        } 
    } 
}

// code beigns in main
int main()
{


/*I'll assume you already know that we should process the children from the biggest Ai first.
Define dp[l][r] as the maximum happiness we can get if we want to put the remaining children to index l to r. 
The answer will be dp[1][N]. 
If the current state is (l, r), that means we are currently processing child with index i = N — (r — l + 1) + 1. 
We will have two choices, either put this child to index l or index r.
If we put it to index l, then the happiness will be abs(l — child_initial_position) * A + dp[l + 1][r]. 
If we put it to index r, then the happiness will be abs(r — child_initial_position) * A + dp[l][r — 1].
We will take the maximum of those. Hope it helps.*/
    
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    ll t=1;
    // cin >> t;
    
    while(t--)
    {
        ll n;
        cin >> n;

        vp v;
        rep(i,0,n)
        {
            ll val;
            cin >> val;
            v.pb({val,i+1});
        }

        sort(v.begin(),v.end(),greater<>());

        ll dp[n+2][n+2];
        clr(dp);

        rep(i,1,n+1)
        {
            repb(j,i,0)
            {
                ll cnt=j-i+n-1;
                dp[j][i]=max(dp[j+1][i]+v[cnt].F*abs(v[cnt].S-j),dp[j][i-1]+v[cnt].F*abs(v[cnt].S-i));
            }
        }
        
        cout << dp[1][n] << endl;
    }
}



Measured memory_bytes: 35048