#include<stdio.h>
int a[32];
long long sol(int l,int r,long long xl,long long xr)
{
	if(r==l+1)
	{
		return a[l]*xl+a[r]*xr;
	}
	long long ans=1000000000000000000LL;
	for(int i=l+1;i<r;i++)
	{
		long long ans2=sol(l,i,xl,xl+xr)+sol(i,r,xl+xr,xr)-a[i]*(xl+xr);
		if(ans2<ans)
		{
			ans=ans2;
		}
	}
	return ans;
}
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
	}
	printf("%lld\n",sol(0,n-1,1,1));
	return 0;
}
### 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>
int a[32];
long long sol(int l,int r,long long xl,long long xr)
{
	if(r==l+1)
	{
		return a[l]*xl+a[r]*xr;
	}
	long long ans=1000000000000000000LL;
	for(int i=l+1;i<r;i++)
	{
		long long ans2=sol(l,i,xl,xl+xr)+sol(i,r,xl+xr,xr)-a[i]*(xl+xr);
		if(ans2<ans)
		{
			ans=ans2;
		}
	}
	return ans;
}
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
	}
	printf("%lld\n",sol(0,n-1,1,1));
	return 0;
}
```