#include<algorithm>
#include<cstdio>
#define LL long long
using namespace std;
int n,sl,fh,a[20];
int rd()
{
	sl=0;fh=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch) {if(ch=='-') fh=-1; ch=getchar();}
	while('0'<=ch&&ch<='9') sl=sl*10+ch-'0',ch=getchar();
	return sl*fh;
}
void chkmin(LL &x,LL y) {if(y<x) x=y;}
LL dfs(int l,int r,int x,int y)
{
	if(r-l<=1) return 0;
	LL res=1e18;
	for(int i=l+1;i<r;++i)
		chkmin(res,dfs(l,i,x,x+y)+dfs(i,r,x+y,y)+1ll*a[i]*(x+y));
	return res;
}
int main()
{
	n=rd();
	for(int i=1;i<=n;++i) a[i]=rd();
	printf("%lld\n",a[1]+a[n]+dfs(1,n,1,1));
	return 0;
}
### Prompt

In cpp, your task is to solve 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<algorithm>
#include<cstdio>
#define LL long long
using namespace std;
int n,sl,fh,a[20];
int rd()
{
	sl=0;fh=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch) {if(ch=='-') fh=-1; ch=getchar();}
	while('0'<=ch&&ch<='9') sl=sl*10+ch-'0',ch=getchar();
	return sl*fh;
}
void chkmin(LL &x,LL y) {if(y<x) x=y;}
LL dfs(int l,int r,int x,int y)
{
	if(r-l<=1) return 0;
	LL res=1e18;
	for(int i=l+1;i<r;++i)
		chkmin(res,dfs(l,i,x,x+y)+dfs(i,r,x+y,y)+1ll*a[i]*(x+y));
	return res;
}
int main()
{
	n=rd();
	for(int i=1;i<=n;++i) a[i]=rd();
	printf("%lld\n",a[1]+a[n]+dfs(1,n,1,1));
	return 0;
}
```