#include<bits/stdc++.h>
typedef long long ll;
const int N=19;
int n,a[N],i;
ll ans,f[66000][N][N];
ll solve(int x,int l,int r,int u,int v){
	if(l==r)return 1ll*(u+v)*a[l];
	ll&ans=f[x][l][r];
	if(ans<1ll<<40)return ans;
	for(int i=l;i<=r;++i){
		ll nw=1ll*(u+v)*a[i];
		if(l<i)nw+=solve(x<<1,l,i-1,u,u+v);
		if(i<r)nw+=solve(x<<1|1,i+1,r,u+v,v);
		if(nw<ans)ans=nw;
	}
	return ans;
}
int main(){
	scanf("%d",&n);
	for(i=1;i<=n;++i)scanf("%d",a+i);
	ans=a[1]+a[n];
	memset(f,1,sizeof f);
	if(n>2)ans+=solve(1,2,n-1,1,1);
	printf("%lld\n",ans);
	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>
typedef long long ll;
const int N=19;
int n,a[N],i;
ll ans,f[66000][N][N];
ll solve(int x,int l,int r,int u,int v){
	if(l==r)return 1ll*(u+v)*a[l];
	ll&ans=f[x][l][r];
	if(ans<1ll<<40)return ans;
	for(int i=l;i<=r;++i){
		ll nw=1ll*(u+v)*a[i];
		if(l<i)nw+=solve(x<<1,l,i-1,u,u+v);
		if(i<r)nw+=solve(x<<1|1,i+1,r,u+v,v);
		if(nw<ans)ans=nw;
	}
	return ans;
}
int main(){
	scanf("%d",&n);
	for(i=1;i<=n;++i)scanf("%d",a+i);
	ans=a[1]+a[n];
	memset(f,1,sizeof f);
	if(n>2)ans+=solve(1,2,n-1,1,1);
	printf("%lld\n",ans);
	return 0;
}
```