#include <bits/stdc++.h>
using namespace std;

const int Maxn = 20;
int n, a[Maxn];
long long work(int lt, int rt, int val1, int val2)
{
	if (lt + 1 == rt) return 0;
	long long tmp = 0x3f3f3f3f3f3f3f3fLL;
	for (int i = lt + 1; i < rt; i++)
		tmp = min(tmp, work(lt, i, val1, val1 + val2) + work(i, rt, val1 + val2, val2) + a[i] * (long long) (val1 + val2));
	return tmp;
}
int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d", &a[i]);
	printf("%lld\n", work(1, n, 1, 1) + a[1] + a[n]);
	return 0;
}
### Prompt

Your task is to create 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>
using namespace std;

const int Maxn = 20;
int n, a[Maxn];
long long work(int lt, int rt, int val1, int val2)
{
	if (lt + 1 == rt) return 0;
	long long tmp = 0x3f3f3f3f3f3f3f3fLL;
	for (int i = lt + 1; i < rt; i++)
		tmp = min(tmp, work(lt, i, val1, val1 + val2) + work(i, rt, val1 + val2, val2) + a[i] * (long long) (val1 + val2));
	return tmp;
}
int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d", &a[i]);
	printf("%lld\n", work(1, n, 1, 1) + a[1] + a[n]);
	return 0;
}
```