/*Lucky_Glass*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N=18;
typedef long long ll;

int n;
ll A[N+3];

ll DFS(int Pl,int Pr,ll Vl,ll Vr){
	int siz=Pr-Pl+1;
	if(siz==2) return 0;
	ll ans=(1ll<<62);
	for(int i=Pl+1;i<=Pr-1;i++)
		ans=min(ans,DFS(Pl,i,Vl,Vl+Vr)+DFS(i,Pr,Vl+Vr,Vr)+(Vl+Vr)*A[i]);
	return ans;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%lld",&A[i]);
	printf("%lld\n",DFS(1,n,1,1)+A[1]+A[n]);
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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
/*Lucky_Glass*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N=18;
typedef long long ll;

int n;
ll A[N+3];

ll DFS(int Pl,int Pr,ll Vl,ll Vr){
	int siz=Pr-Pl+1;
	if(siz==2) return 0;
	ll ans=(1ll<<62);
	for(int i=Pl+1;i<=Pr-1;i++)
		ans=min(ans,DFS(Pl,i,Vl,Vl+Vr)+DFS(i,Pr,Vl+Vr,Vr)+(Vl+Vr)*A[i]);
	return ans;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%lld",&A[i]);
	printf("%lld\n",DFS(1,n,1,1)+A[1]+A[n]);
	return 0;
}
```