#include<cstdio>
#include<iostream>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;
const int N=20;
map<pair<int,int>,ll> f[N][N];
ll a[N],n;
ll dp(int l,int r,int x,int y){
	if (l>r)return 0;
	if (l==r)return a[l]*(x+y);
	if (f[l][r].count(make_pair(x,y)))
		return f[l][r][make_pair(x,y)];
	ll ans=1e18;
	for (int i=l;i<=r;i++)
		ans=min(ans,a[i]*(x+y)+dp(l,i-1,x,x+y)+dp(i+1,r,x+y,y));
	return f[l][r][make_pair(x,y)]=ans;
}
int main(){
	cin>>n;n-=2;
	for (int i=0;i<=n+1;i++)scanf("%lld",&a[i]);
	cout<<dp(1,n,1,1)+a[0]+a[n+1]<<endl;
}
### Prompt

Create a solution in CPP for 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<cstdio>
#include<iostream>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;
const int N=20;
map<pair<int,int>,ll> f[N][N];
ll a[N],n;
ll dp(int l,int r,int x,int y){
	if (l>r)return 0;
	if (l==r)return a[l]*(x+y);
	if (f[l][r].count(make_pair(x,y)))
		return f[l][r][make_pair(x,y)];
	ll ans=1e18;
	for (int i=l;i<=r;i++)
		ans=min(ans,a[i]*(x+y)+dp(l,i-1,x,x+y)+dp(i+1,r,x+y,y));
	return f[l][r][make_pair(x,y)]=ans;
}
int main(){
	cin>>n;n-=2;
	for (int i=0;i<=n+1;i++)scanf("%lld",&a[i]);
	cout<<dp(1,n,1,1)+a[0]+a[n+1]<<endl;
}
```