#include<cstdio>
#include<algorithm>
#include<map>
#define ll long long
#define MP make_pair
#define Pi pair<int,int>
#define Pl pair<ll,ll>
#define PP pair<Pi,Pl >
using namespace std;
#define N 20
#define inf 100000000000000ll
int n;
ll num[N];
map<PP,ll>p;
ll work(int l,int r,ll x,ll y)
{
	if(r-l<=1) return 0;
	PP u=MP(MP(l,r),MP(x,y));
	if(p.count(u)) return p[u];
	ll res=inf;
	for(int i=l+1;i<=r-1;i++)
	res=min(res,work(l,i,x,x+y)+work(i,r,x+y,y)+num[i]*(x+y));
	return p[u]=res;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%lld",&num[i]);
	printf("%lld\n",num[1]+num[n]+work(1,n,1,1));
}
### Prompt

Please create a solution in Cpp 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>
#include<map>
#define ll long long
#define MP make_pair
#define Pi pair<int,int>
#define Pl pair<ll,ll>
#define PP pair<Pi,Pl >
using namespace std;
#define N 20
#define inf 100000000000000ll
int n;
ll num[N];
map<PP,ll>p;
ll work(int l,int r,ll x,ll y)
{
	if(r-l<=1) return 0;
	PP u=MP(MP(l,r),MP(x,y));
	if(p.count(u)) return p[u];
	ll res=inf;
	for(int i=l+1;i<=r-1;i++)
	res=min(res,work(l,i,x,x+y)+work(i,r,x+y,y)+num[i]*(x+y));
	return p[u]=res;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%lld",&num[i]);
	printf("%lld\n",num[1]+num[n]+work(1,n,1,1));
}
```