#include<bits/stdc++.h>
#define MAXN 19
#define INF 9223372036854775807
using namespace std;

inline long long read()
{
	long long x=0,f=1;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-') f=-f;ch=getchar();}
	while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

long long n,m;
long long a[MAXN];
long long dfs(long long l,long long r,long long x,long long y)
{
	if(l==r-1) return a[l]*x+a[r]*y;
	long long ans=INF;
	for(register long long m=l+1;m<r;++m) ans=min(ans,dfs(l,m,x,x+y)+dfs(m,r,x+y,y)-a[m]*(x+y));
	return ans;
}

int main()
{
	n=read();
	for(register long long i=1;i<=n;++i) a[i]=read();
	printf("%lld",dfs(1,n,1,1));
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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>
#define MAXN 19
#define INF 9223372036854775807
using namespace std;

inline long long read()
{
	long long x=0,f=1;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-') f=-f;ch=getchar();}
	while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

long long n,m;
long long a[MAXN];
long long dfs(long long l,long long r,long long x,long long y)
{
	if(l==r-1) return a[l]*x+a[r]*y;
	long long ans=INF;
	for(register long long m=l+1;m<r;++m) ans=min(ans,dfs(l,m,x,x+y)+dfs(m,r,x+y,y)-a[m]*(x+y));
	return ans;
}

int main()
{
	n=read();
	for(register long long i=1;i<=n;++i) a[i]=read();
	printf("%lld",dfs(1,n,1,1));
	return 0;
}
```