#include <iostream>
using namespace std;

long long a[20];

long long dfs(int l, int r, int lx, int rx){
    if(l + 1 == r){
        return 0;
    }
    long long res = 200000000000000;
    for(int k = l+1; k < r; k++){
        res = min(res, dfs(l, k, lx, lx + rx) + dfs(k, r, lx + rx, rx) + a[k]*(lx+rx));
    }
    return res;
}

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

Construct a Cpp code solution to the problem outlined:

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 <iostream>
using namespace std;

long long a[20];

long long dfs(int l, int r, int lx, int rx){
    if(l + 1 == r){
        return 0;
    }
    long long res = 200000000000000;
    for(int k = l+1; k < r; k++){
        res = min(res, dfs(l, k, lx, lx + rx) + dfs(k, r, lx + rx, rx) + a[k]*(lx+rx));
    }
    return res;
}

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