#include <bits/stdc++.h>
using namespace std;
#define lor(a,b,c) for(register int a=b;a<=c;++a)
#define ror(a,b,c) for(register int a=c;a>=b;--a)
typedef long long ll;

const int N=20;
const ll INF=1e18;

int n; ll a[N];

ll dfs(int l,int r,ll x,ll y){
	if(r-l<=1) return a[l]*x+a[r]*y;
	ll ans=INF;
	lor(i,l+1,r-1) ans=min(ans,dfs(l,i,x,x+y)+dfs(i,r,x+y,y)-a[i]*(x+y));
	return ans;
}

int main(){
	scanf("%d",&n); lor(i,1,n) scanf("%lld",&a[i]);
	printf("%lld\n",dfs(1,n,1,1));

	return 0;
}
### Prompt

Please create a solution in Cpp 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>
using namespace std;
#define lor(a,b,c) for(register int a=b;a<=c;++a)
#define ror(a,b,c) for(register int a=c;a>=b;--a)
typedef long long ll;

const int N=20;
const ll INF=1e18;

int n; ll a[N];

ll dfs(int l,int r,ll x,ll y){
	if(r-l<=1) return a[l]*x+a[r]*y;
	ll ans=INF;
	lor(i,l+1,r-1) ans=min(ans,dfs(l,i,x,x+y)+dfs(i,r,x+y,y)-a[i]*(x+y));
	return ans;
}

int main(){
	scanf("%d",&n); lor(i,1,n) scanf("%lld",&a[i]);
	printf("%lld\n",dfs(1,n,1,1));

	return 0;
}
```