#include<bits/stdc++.h>
const int N=18;
using namespace std;
typedef long long ll;
const ll inf=1e18;
int n,a[N],x,y,nx,ny,di;
ll f[N][N][1<<N],pl[N],pr[N],now;
int main(){
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	scanf("%d",a+i),pl[i]=pr[i]=inf;
	pl[0]=pr[n-1]=0;
	for(int s=(1<<n-1)-1;s;s--){
		x=y=1,nx=ny=0;
		for(di=1;di<<1<=s;di<<=1)
		if(s&di)ny++,y+=x;else nx++,x+=y;
		for(int i=nx;i<n;i++)
		for(int j=i+2;j+ny<n;j++){
			now=inf;
			for(int k=i+1;k<j;k++)
			now=min(now,f[i][k][s^(di<<1)]+
			f[k][j][s^di^(di<<1)]+a[k]*1ll*(x+y));
			f[i][j][s]=now;
		}
	}
	printf("%lld",f[0][n-1][1]+a[0]+a[n-1]);
}
### Prompt

Your task is to create 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<bits/stdc++.h>
const int N=18;
using namespace std;
typedef long long ll;
const ll inf=1e18;
int n,a[N],x,y,nx,ny,di;
ll f[N][N][1<<N],pl[N],pr[N],now;
int main(){
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	scanf("%d",a+i),pl[i]=pr[i]=inf;
	pl[0]=pr[n-1]=0;
	for(int s=(1<<n-1)-1;s;s--){
		x=y=1,nx=ny=0;
		for(di=1;di<<1<=s;di<<=1)
		if(s&di)ny++,y+=x;else nx++,x+=y;
		for(int i=nx;i<n;i++)
		for(int j=i+2;j+ny<n;j++){
			now=inf;
			for(int k=i+1;k<j;k++)
			now=min(now,f[i][k][s^(di<<1)]+
			f[k][j][s^di^(di<<1)]+a[k]*1ll*(x+y));
			f[i][j][s]=now;
		}
	}
	printf("%lld",f[0][n-1][1]+a[0]+a[n-1]);
}
```