#include<iostream>
#include<cstdio>
#include<map>
using namespace std;
#define N 19
#define LL long long
#define INF 0x3f3f3f3f3f3f3f3f
int n,a[N];map<pair<int,int>,LL>f[N][N];
LL dfs(int l,int r,int v1,int v2)
{
	if(r-l<=1)return 0;pair<int,int>p1=make_pair(v1,v2);if(f[l][r].count(p1))return f[l][r][p1];LL &v=f[l][r][p1];v=INF;
	for(int i=l+1;i<r;i++)v=min(v,dfs(l,i,v1,v1+v2)+dfs(i,r,v1+v2,v2)+1ll*(v1+v2)*a[i]);return v;
}
int main(){
	scanf("%d",&n);int i;
	for(i=1;i<=n;i++)scanf("%d",&a[i]);printf("%lld",dfs(1,n,1,1)+a[1]+a[n]);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<cstdio>
#include<map>
using namespace std;
#define N 19
#define LL long long
#define INF 0x3f3f3f3f3f3f3f3f
int n,a[N];map<pair<int,int>,LL>f[N][N];
LL dfs(int l,int r,int v1,int v2)
{
	if(r-l<=1)return 0;pair<int,int>p1=make_pair(v1,v2);if(f[l][r].count(p1))return f[l][r][p1];LL &v=f[l][r][p1];v=INF;
	for(int i=l+1;i<r;i++)v=min(v,dfs(l,i,v1,v1+v2)+dfs(i,r,v1+v2,v2)+1ll*(v1+v2)*a[i]);return v;
}
int main(){
	scanf("%d",&n);int i;
	for(i=1;i<=n;i++)scanf("%d",&a[i]);printf("%lld",dfs(1,n,1,1)+a[1]+a[n]);return 0;
}
```