#include <stdio.h>
#include <iostream>
using namespace std;

int N;
int A[18];

long long rec(int l, int r, long long xl, long long xr) {
    if (r - l == 1) {
        return 0;
    }
    
    long long ret = (long long) 1e18;
    for (int i = l + 1; i < r; i++) {
        long long tmp = A[i] * (xl+xr) + rec(l, i, xl, xl+xr) + rec(i, r, xl+xr, xr);
        ret = min(ret, tmp);
    }
    return ret;
}

int main() {    
    cin >> N;
    for (int i = 0; i < N; i++) {
        cin >> A[i];
    }
    
    cout << rec(0, N - 1, 1, 1) + A[0] + A[N - 1];
    
}
### Prompt

In cpp, your task is to solve 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 <stdio.h>
#include <iostream>
using namespace std;

int N;
int A[18];

long long rec(int l, int r, long long xl, long long xr) {
    if (r - l == 1) {
        return 0;
    }
    
    long long ret = (long long) 1e18;
    for (int i = l + 1; i < r; i++) {
        long long tmp = A[i] * (xl+xr) + rec(l, i, xl, xl+xr) + rec(i, r, xl+xr, xr);
        ret = min(ret, tmp);
    }
    return ret;
}

int main() {    
    cin >> N;
    for (int i = 0; i < N; i++) {
        cin >> A[i];
    }
    
    cout << rec(0, N - 1, 1, 1) + A[0] + A[N - 1];
    
}
```