#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=20;
long long inf;
long long a[N];
long long dp(int l,int r,long long xl,long long xr)
{
	//printf("%d,%d,%lld,%lld\n",l,r,xl,xr);
	if(l+1==r)
		return 0;
	long long ans=inf;
	//printf("%d,%d,%lld,%lld <%lld>\n",l,r,xl,xr,ans);
	int i;
	for(i=l+1;i<r;i++)
		ans=min(ans,dp(l,i,xl,xl+xr)+dp(i,r,xl+xr,xr)+(xl+xr)*a[i]);
	//printf("%d,%d,%lld,%lld  %lld\n",l,r,xl,xr,ans);
	return ans;
}
int main()
{
	inf=1<<30;
	inf<<=30;
	int n,i;
	long long ans;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
		cin>>a[i];
	ans=dp(1,n,1,1);
	ans+=a[1]+a[n];
	cout<<ans;
	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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=20;
long long inf;
long long a[N];
long long dp(int l,int r,long long xl,long long xr)
{
	//printf("%d,%d,%lld,%lld\n",l,r,xl,xr);
	if(l+1==r)
		return 0;
	long long ans=inf;
	//printf("%d,%d,%lld,%lld <%lld>\n",l,r,xl,xr,ans);
	int i;
	for(i=l+1;i<r;i++)
		ans=min(ans,dp(l,i,xl,xl+xr)+dp(i,r,xl+xr,xr)+(xl+xr)*a[i]);
	//printf("%d,%d,%lld,%lld  %lld\n",l,r,xl,xr,ans);
	return ans;
}
int main()
{
	inf=1<<30;
	inf<<=30;
	int n,i;
	long long ans;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
		cin>>a[i];
	ans=dp(1,n,1,1);
	ans+=a[1]+a[n];
	cout<<ans;
	return 0;
}
```