#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#define N 25
#define ll long long
using namespace std;
int n;
int s[N];
ll find(int li,int ri,int x,int y){
	if(li>=ri-1) return 0;
	ll mi=1e17;
	for(int i=li+1;i<ri;i++) mi=min(mi,find(li,i,x,x+y)+find(i,ri,x+y,y)+(ll)s[i]*(x+y));
	return mi;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&s[i]);
	printf("%lld\n",find(1,n,1,1)+s[1]+s[n]);
	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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#define N 25
#define ll long long
using namespace std;
int n;
int s[N];
ll find(int li,int ri,int x,int y){
	if(li>=ri-1) return 0;
	ll mi=1e17;
	for(int i=li+1;i<ri;i++) mi=min(mi,find(li,i,x,x+y)+find(i,ri,x+y,y)+(ll)s[i]*(x+y));
	return mi;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&s[i]);
	printf("%lld\n",find(1,n,1,1)+s[1]+s[n]);
	return 0;
} 
```