#include <bits/stdc++.h>

typedef long long ll;

const ll INF = 1e18;

using namespace std;

ll n, a[20];

ll calc(int l, int xl, int r, int xr)
{
    ll ans = INF;
    if (l + 1 == r)
    {
        return 0;
    }
    int xk = xl + xr;
    for (int k = l + 1; k < r; k++)
    {
        ans = min(ans, calc(l, xl, k, xk) + a[k] * xk + calc(k, xk, r, xr));
    }
    return ans;
}

int main()
{
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    cout << a[0] + calc(0, 1, n - 1, 1) + a[n - 1];
}

### Prompt

Generate a Cpp solution to the following problem:

There is a stack of N cards, each of which has a non-negative integer written on it. The integer written on the i-th card from the top is A_i.

Snuke will repeat the following operation until two cards remain:

* Choose three consecutive cards from the stack.
* Eat the middle card of the three.
* For each of the other two cards, replace the integer written on it by the sum of that integer and the integer written on the card eaten.
* Return the two cards to the original position in the stack, without swapping them.



Find the minimum possible sum of the integers written on the last two cards remaining.

Constraints

* 2 \leq N \leq 18
* 0 \leq A_i \leq 10^9 (1\leq i\leq N)
* 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 minimum possible sum of the integers written on the last two cards remaining.

Examples

Input

4
3 1 4 2


Output

16


Input

6
5 2 4 1 6 9


Output

51


Input

10
3 1 4 1 5 9 2 6 5 3


Output

115

### Response

```cpp
#include <bits/stdc++.h>

typedef long long ll;

const ll INF = 1e18;

using namespace std;

ll n, a[20];

ll calc(int l, int xl, int r, int xr)
{
    ll ans = INF;
    if (l + 1 == r)
    {
        return 0;
    }
    int xk = xl + xr;
    for (int k = l + 1; k < r; k++)
    {
        ans = min(ans, calc(l, xl, k, xk) + a[k] * xk + calc(k, xk, r, xr));
    }
    return ans;
}

int main()
{
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    cout << a[0] + calc(0, 1, n - 1, 1) + a[n - 1];
}

```