#include <iostream>
#include <algorithm>
#include <string>
#include <math.h>
#include <vector>
using namespace std;
typedef long long int ll;
int n;
std::vector<ll> a;
ll dfs(int l,int r,ll A,ll b){
	if(l>r)return 0;
	ll ans=1e18;
	for(int i=l;i<=r;i++){
		ans=min(ans,a[i]*(A+b)+dfs(l,i-1,A,A+b)+dfs(i+1,r,A+b,b));
	}
	return ans;
}

int main(){
	cin >> n;
	a=vector<ll> (n);
	for(int i=0;i<n;i++){
		cin >> a[i];
	}
	cout << a[0]+a[n-1]+dfs(1,n-2,1,1) << endl;
	return 0;
}
### Prompt

Generate 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 <iostream>
#include <algorithm>
#include <string>
#include <math.h>
#include <vector>
using namespace std;
typedef long long int ll;
int n;
std::vector<ll> a;
ll dfs(int l,int r,ll A,ll b){
	if(l>r)return 0;
	ll ans=1e18;
	for(int i=l;i<=r;i++){
		ans=min(ans,a[i]*(A+b)+dfs(l,i-1,A,A+b)+dfs(i+1,r,A+b,b));
	}
	return ans;
}

int main(){
	cin >> n;
	a=vector<ll> (n);
	for(int i=0;i<n;i++){
		cin >> a[i];
	}
	cout << a[0]+a[n-1]+dfs(1,n-2,1,1) << endl;
	return 0;
}
```