#include<map>
#include<cstdio>
#include<algorithm>
using namespace std;
#define LL long long
#define pr(a,b) make_pair(a,b)
const int MAXN=20;
const long long INF=5e18;
int N;long long A[MAXN];
long long Solve(int L,int R,LL X,LL Y){
	if(L+1>=R)return 0;
	long long ret=INF;
	for(int i=L+1;i<=R-1;i++)
		ret=min(ret,Solve(L,i,X,X+Y)+Solve(i,R,X+Y,Y)+A[i]*(X+Y));
	return ret;
}
int main(){
	scanf("%d",&N);
	for(int i=1;i<=N;i++)
		scanf("%lld",&A[i]);
	printf("%lld\n",A[1]+A[N]+Solve(1,N,1,1));
}
### Prompt

Create a solution in cpp for 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<map>
#include<cstdio>
#include<algorithm>
using namespace std;
#define LL long long
#define pr(a,b) make_pair(a,b)
const int MAXN=20;
const long long INF=5e18;
int N;long long A[MAXN];
long long Solve(int L,int R,LL X,LL Y){
	if(L+1>=R)return 0;
	long long ret=INF;
	for(int i=L+1;i<=R-1;i++)
		ret=min(ret,Solve(L,i,X,X+Y)+Solve(i,R,X+Y,Y)+A[i]*(X+Y));
	return ret;
}
int main(){
	scanf("%d",&N);
	for(int i=1;i<=N;i++)
		scanf("%lld",&A[i]);
	printf("%lld\n",A[1]+A[N]+Solve(1,N,1,1));
}
```