#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long f[20][20][1<<18];
int l[1<<18],r[1<<18],a[19];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	l[1]=r[1]=1;
	for(int i=2;i<(1<<n);i++){
		if(i&1)l[i]=l[i>>1]+r[i>>1],r[i]=r[i>>1];
		else l[i]=l[i>>1],r[i]=l[i>>1]+r[i>>1];
	}
	memset(f,63,sizeof(f));
	for(int i=1;i<n;i++)
		for(int j=1;j<(1<<n);j++)f[i][i+1][j]=0;
	for(int L=2;L<n;L++)
		for(int i=1;i<=n-L;i++){
			int j=i+L;
			for(int k=i+1;k<j;k++)
				for(int p=1;p<(1<<(n-L));p++)f[i][j][p]=min(f[i][j][p],f[i][k][p<<1]+f[k][j][(p<<1)|1]+1ll*(l[p]+r[p])*a[k]);
		}
	printf("%lld\n",f[1][n][1]+a[1]+a[n]);
	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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long f[20][20][1<<18];
int l[1<<18],r[1<<18],a[19];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	l[1]=r[1]=1;
	for(int i=2;i<(1<<n);i++){
		if(i&1)l[i]=l[i>>1]+r[i>>1],r[i]=r[i>>1];
		else l[i]=l[i>>1],r[i]=l[i>>1]+r[i>>1];
	}
	memset(f,63,sizeof(f));
	for(int i=1;i<n;i++)
		for(int j=1;j<(1<<n);j++)f[i][i+1][j]=0;
	for(int L=2;L<n;L++)
		for(int i=1;i<=n-L;i++){
			int j=i+L;
			for(int k=i+1;k<j;k++)
				for(int p=1;p<(1<<(n-L));p++)f[i][j][p]=min(f[i][j][p],f[i][k][p<<1]+f[k][j][(p<<1)|1]+1ll*(l[p]+r[p])*a[k]);
		}
	printf("%lld\n",f[1][n][1]+a[1]+a[n]);
	return 0;
}
```