#include<iostream>
#include<cstdio>
#define FOR(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
typedef long long ll;
const ll INF=1e18;
int n,a[100];
ll dfs(int l,int r,int x,int y){
	if(l>r) return 0;ll ret=INF;
	FOR(i,l,r) ret=min(ret,dfs(l,i-1,x,x+y)+dfs(i+1,r,x+y,y)+1ll*(x+y)*a[i]);
	return ret;
}
int main(){
	scanf("%d",&n);
	FOR(i,1,n) scanf("%d",&a[i]);
	cout<<dfs(2,n-1,1,1)+a[1]+a[n]<<'\n';
}
### Prompt

Your task is to create 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<cstdio>
#define FOR(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
typedef long long ll;
const ll INF=1e18;
int n,a[100];
ll dfs(int l,int r,int x,int y){
	if(l>r) return 0;ll ret=INF;
	FOR(i,l,r) ret=min(ret,dfs(l,i-1,x,x+y)+dfs(i+1,r,x+y,y)+1ll*(x+y)*a[i]);
	return ret;
}
int main(){
	scanf("%d",&n);
	FOR(i,1,n) scanf("%d",&a[i]);
	cout<<dfs(2,n-1,1,1)+a[1]+a[n]<<'\n';
}
```