#include <bits/stdc++.h>
#define FR first
#define SE second
#define inf 0x3f3f3f3f3f3f3f3fLL

using namespace std;

typedef long long ll;
typedef pair<ll,ll> pr;

map <pr,ll> f[20][20];
int num[20];

ll dfs(int l,int r,ll s1,ll s2) {
  if (r-l==1) return s1*num[l]+s2*num[r];
  if (f[l][r].count(pr(s1,s2))) return f[l][r][pr(s1,s2)];
  ll ans=inf;
  for(int i=l+1;i<r;i++)
    ans=min(ans,dfs(l,i,s1,s1+s2)+dfs(i,r,s1+s2,s2)-num[i]*(s1+s2));
  return f[l][r][pr(s1,s2)]=ans;
}

int main() {
  int n;
  scanf("%d",&n);
  for(int i=1;i<=n;i++) scanf("%d",&num[i]);
  printf("%lld\n",dfs(1,n,1,1));
  return 0;
}
### Prompt

Your challenge is to write a Cpp solution 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 <bits/stdc++.h>
#define FR first
#define SE second
#define inf 0x3f3f3f3f3f3f3f3fLL

using namespace std;

typedef long long ll;
typedef pair<ll,ll> pr;

map <pr,ll> f[20][20];
int num[20];

ll dfs(int l,int r,ll s1,ll s2) {
  if (r-l==1) return s1*num[l]+s2*num[r];
  if (f[l][r].count(pr(s1,s2))) return f[l][r][pr(s1,s2)];
  ll ans=inf;
  for(int i=l+1;i<r;i++)
    ans=min(ans,dfs(l,i,s1,s1+s2)+dfs(i,r,s1+s2,s2)-num[i]*(s1+s2));
  return f[l][r][pr(s1,s2)]=ans;
}

int main() {
  int n;
  scanf("%d",&n);
  for(int i=1;i<=n;i++) scanf("%d",&num[i]);
  printf("%lld\n",dfs(1,n,1,1));
  return 0;
}
```