#include <cstdio>
#include <algorithm>

using namespace std;

int a[18];

long long dfs(int x, int y, int p, int q) {
    int i;
    long long ans = 1e18;
    
    if (x + 1 == y) return 0;
    
    for (i = x + 1; i < y; i++) ans = min(ans, dfs(x, i, p, p + q) + dfs(i, y, p + q, q) + (long long)(p + q) * a[i]);
    
    return ans;
}

int main() {
    int n, i;
    
    scanf("%d", &n);
    
    for (i = 0; i < n; i++) scanf("%d", &a[i]);
    
    printf("%lld\n", dfs(0, n - 1, 1, 1) + a[0] + a[n - 1]);
    
    return 0;
}

### Prompt

Please formulate 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 <cstdio>
#include <algorithm>

using namespace std;

int a[18];

long long dfs(int x, int y, int p, int q) {
    int i;
    long long ans = 1e18;
    
    if (x + 1 == y) return 0;
    
    for (i = x + 1; i < y; i++) ans = min(ans, dfs(x, i, p, p + q) + dfs(i, y, p + q, q) + (long long)(p + q) * a[i]);
    
    return ans;
}

int main() {
    int n, i;
    
    scanf("%d", &n);
    
    for (i = 0; i < n; i++) scanf("%d", &a[i]);
    
    printf("%lld\n", dfs(0, n - 1, 1, 1) + a[0] + a[n - 1]);
    
    return 0;
}

```