#include<cstdio>
#include<algorithm>
typedef long long LL;
int A[20],n;
LL f(int l,int r,int x,int y){
    if(r-l<=1)return 0;
    LL p=1e18;
    for(int i=l+1;i<r;++i)
    p=std::min(p,f(l,i,x,x+y)+f(i,r,x+y,y)+(LL)(x+y)*A[i]);
    return p;
}
int main(){
    scanf("%d",&n);
    for(int i=0;i<n;++i)scanf("%d",A+i);
    printf("%lld",f(0,n-1,1,1)+*A+A[n-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<cstdio>
#include<algorithm>
typedef long long LL;
int A[20],n;
LL f(int l,int r,int x,int y){
    if(r-l<=1)return 0;
    LL p=1e18;
    for(int i=l+1;i<r;++i)
    p=std::min(p,f(l,i,x,x+y)+f(i,r,x+y,y)+(LL)(x+y)*A[i]);
    return p;
}
int main(){
    scanf("%d",&n);
    for(int i=0;i<n;++i)scanf("%d",A+i);
    printf("%lld",f(0,n-1,1,1)+*A+A[n-1]);
    return 0;
}
```