The Fair Nut lives in n story house. a_i people live on the i-th floor of the house. Every person uses elevator twice a day: to get from the floor where he/she lives to the ground (first) floor and to get from the first floor to the floor where he/she lives, when he/she comes back home in the evening. 

It was decided that elevator, when it is not used, will stay on the x-th floor, but x hasn't been chosen yet. When a person needs to get from floor a to floor b, elevator follows the simple algorithm: 

  * Moves from the x-th floor (initially it stays on the x-th floor) to the a-th and takes the passenger. 
  * Moves from the a-th floor to the b-th floor and lets out the passenger (if a equals b, elevator just opens and closes the doors, but still comes to the floor from the x-th floor). 
  * Moves from the b-th floor back to the x-th. 

The elevator never transposes more than one person and always goes back to the floor x before transposing a next passenger. The elevator spends one unit of electricity to move between neighboring floors. So moving from the a-th floor to the b-th floor requires |a - b| units of electricity.

Your task is to help Nut to find the minimum number of electricity units, that it would be enough for one day, by choosing an optimal the x-th floor. Don't forget than elevator initially stays on the x-th floor. 

Input

The first line contains one integer n (1 ≤ n ≤ 100) — the number of floors.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 100) — the number of people on each floor.

Output

In a single line, print the answer to the problem — the minimum number of electricity units.

Examples

Input

3
0 2 1


Output

16

Input

2
1 1


Output

4

Note

In the first example, the answer can be achieved by choosing the second floor as the x-th floor. Each person from the second floor (there are two of them) would spend 4 units of electricity per day (2 to get down and 2 to get up), and one person from the third would spend 8 units of electricity per day (4 to get down and 4 to get up). 4 ⋅ 2 + 8 ⋅ 1 = 16.

In the second example, the answer can be achieved by choosing the first floor as the x-th floor.
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update
using namespace __gnu_pbds;
#define int long long
#define pii pair<int,int>
#define ordered_set tree<pii, null_type,less<pii>, rb_tree_tag,tree_order_statistics_node_update>
#define fr first
#define se second
#define mp make_pair
const int MOD=(int)1e7+7;
const int inf=(int)1e18;
     
using namespace std;
     
    //custom hash for unordered map
struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }
     
    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};
inline int binexp(int a,int b,int m)
{
    int res=1;
    a%=m;
    while(b)
    {
        if(b&1)
    	    res=(res*a)%m;
        a=(a*a)%m;
        b>>=1;
    }
    return res;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r",stdin);
    freopen("out.txt", "w", stdout);
    #endif
	int n;
	cin >> n;
	int arr[200] = {0};
	for(int i = 1; i <= n; ++i)
		cin >> arr[i];

	int mn = inf;

	for(int i = 2; i <= 2; ++i)
	{
		int t = 0;
		for(int j = 1; j <= n; ++j)
		{
			t += 2 * abs(i - j) * arr[j] + 2 * (i - 1) * arr[j];
			//		cout << t<< "\n";

		}

		mn = min(t, mn);
	//	cout << t<< "\n";
	}
    
	for(int i = 1; i <= n; ++i)
		mn += 2 * (i - 1) * arr[i];

	cout << mn;

    return 0;
}