#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
#include<algorithm>

#define int long long 

using namespace std;

int a[20],n;

int f(int l,int r,int xl,int xr)
{
	int i,j;
	if(r-l<=1) return 0;
	int ans=0x3f3f3f3f3f3f3f3f;
	for(i=l+1;i<r;i++)
	{
		ans=min(ans,f(l,i,xl,xl+xr)+f(i,r,xl+xr,xr)+a[i]*(xl+xr));
	}
	return ans;
}

signed main()
{
	ios::sync_with_stdio(false);
	register int i,j;
	cin>>n;
	for(i=1;i<=n;i++)
	cin>>a[i];
	cout<<a[1]+a[n]+f(1,n,1,1);
}

### 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<cmath>
#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
#include<algorithm>

#define int long long 

using namespace std;

int a[20],n;

int f(int l,int r,int xl,int xr)
{
	int i,j;
	if(r-l<=1) return 0;
	int ans=0x3f3f3f3f3f3f3f3f;
	for(i=l+1;i<r;i++)
	{
		ans=min(ans,f(l,i,xl,xl+xr)+f(i,r,xl+xr,xr)+a[i]*(xl+xr));
	}
	return ans;
}

signed main()
{
	ios::sync_with_stdio(false);
	register int i,j;
	cin>>n;
	for(i=1;i<=n;i++)
	cin>>a[i];
	cout<<a[1]+a[n]+f(1,n,1,1);
}

```