#include <cstdio>
#include <iostream>

using namespace std;

long long n,a[1000010];

long long f(long long l,long long r,long long x,long long y)
{
  long long ans=1e18;
  if(r-l<=1) return 0;
  for(int i=l+1;i<=r-1;i++) ans=min(f(l,i,x,x+y)+f(i,r,x+y,y)+a[i]*(x+y),ans);
  return ans;
}

int main()
{
  scanf("%lld",&n);
  for(int i=1;i<=n;i++)
  {
    scanf("%lld",&a[i]);
  }
  printf("%lld\n",f(1,n,1,1)+a[1]+a[n]);
}

### Prompt

Please provide a CPP coded solution 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
#include <cstdio>
#include <iostream>

using namespace std;

long long n,a[1000010];

long long f(long long l,long long r,long long x,long long y)
{
  long long ans=1e18;
  if(r-l<=1) return 0;
  for(int i=l+1;i<=r-1;i++) ans=min(f(l,i,x,x+y)+f(i,r,x+y,y)+a[i]*(x+y),ans);
  return ans;
}

int main()
{
  scanf("%lld",&n);
  for(int i=1;i<=n;i++)
  {
    scanf("%lld",&a[i]);
  }
  printf("%lld\n",f(1,n,1,1)+a[1]+a[n]);
}

```