#include <iostream>
#include <algorithm>
#include <vector>
#include <tuple>
#include <functional>
#include <limits>
using namespace std;

int main(){
    int N;
    cin>>N;
    vector<int64_t> A(N);
    for(auto &a:A) cin>>a;
    function<int64_t(int64_t,int64_t,int64_t,int64_t)> rec = [&](int64_t l, int64_t r, int64_t lx, int64_t rx){
        if(r-l==1) return int64_t(0);
        int64_t ret = numeric_limits<int64_t>::max();
        for(int k=l+1;k<r;k++){
            ret = min(ret, rec(l,k,lx,lx+rx)+rec(k,r,lx+rx,rx)+A[k]*(lx+rx));
        }
        return ret;
    };
    cout<<rec(0,N-1,1,1)+A[0]+A[N-1]<<endl;
    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 <iostream>
#include <algorithm>
#include <vector>
#include <tuple>
#include <functional>
#include <limits>
using namespace std;

int main(){
    int N;
    cin>>N;
    vector<int64_t> A(N);
    for(auto &a:A) cin>>a;
    function<int64_t(int64_t,int64_t,int64_t,int64_t)> rec = [&](int64_t l, int64_t r, int64_t lx, int64_t rx){
        if(r-l==1) return int64_t(0);
        int64_t ret = numeric_limits<int64_t>::max();
        for(int k=l+1;k<r;k++){
            ret = min(ret, rec(l,k,lx,lx+rx)+rec(k,r,lx+rx,rx)+A[k]*(lx+rx));
        }
        return ret;
    };
    cout<<rec(0,N-1,1,1)+A[0]+A[N-1]<<endl;
    return 0;
}
```