#include<bits/stdc++.h>
using namespace std;
#define int long long
using pint=pair<int,int>;

int N;
int A[18];
map<pair<pint,pint>,int> dp;
int dfs(int l,int r,int kl,int kr){
  if(l+1==r)return 0;
  pair<pint,pint> p={{l,r},{kl,kr}};
  if(dp.find(p)!=dp.end())return dp[p];
  int ret=LLONG_MAX;
  for(int i=l+1;i<r;i++){
    int ki=kl+kr;
    ret=min(ret,ki*A[i]+dfs(l,i,kl,ki)+dfs(i,r,ki,kr));
  }
  return dp[p]=ret;
}
signed main(){
  cin>>N;
  for(int i=0;i<N;i++)cin>>A[i];
  cout<<dfs(0,N-1,1,1)+A[0]+A[N-1]<<endl;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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>
using namespace std;
#define int long long
using pint=pair<int,int>;

int N;
int A[18];
map<pair<pint,pint>,int> dp;
int dfs(int l,int r,int kl,int kr){
  if(l+1==r)return 0;
  pair<pint,pint> p={{l,r},{kl,kr}};
  if(dp.find(p)!=dp.end())return dp[p];
  int ret=LLONG_MAX;
  for(int i=l+1;i<r;i++){
    int ki=kl+kr;
    ret=min(ret,ki*A[i]+dfs(l,i,kl,ki)+dfs(i,r,ki,kr));
  }
  return dp[p]=ret;
}
signed main(){
  cin>>N;
  for(int i=0;i<N;i++)cin>>A[i];
  cout<<dfs(0,N-1,1,1)+A[0]+A[N-1]<<endl;
}
```