#include<iostream>
#include<iomanip>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int read()
{
	char c=getchar();
	int res=0;
	while(c<'0'||c>'9') c=getchar();
	while(c>='0'&&c<='9') res=(res<<1)+(res<<3)+(c^48),c=getchar();
	return res;
}
int n,a[105];
long long dp(int l,int r,int cl,int cr)
{
	if(r-l<=1) return 0;
	long long ans=1e18;
	for(int i=l+1;i<=r-1;i++) ans=min(ans,dp(l,i,cl+cr,cr)+dp(i,r,cl,cr+cl)+1ll*(cl+cr)*a[i]);
	return ans;
}
int main()
{
	n=read();
	for(int i=1;i<=n;i++) a[i]=read();
	printf("%lld\n",dp(1,n,1,1)+a[1]+a[n]);
}
### 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<iostream>
#include<iomanip>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int read()
{
	char c=getchar();
	int res=0;
	while(c<'0'||c>'9') c=getchar();
	while(c>='0'&&c<='9') res=(res<<1)+(res<<3)+(c^48),c=getchar();
	return res;
}
int n,a[105];
long long dp(int l,int r,int cl,int cr)
{
	if(r-l<=1) return 0;
	long long ans=1e18;
	for(int i=l+1;i<=r-1;i++) ans=min(ans,dp(l,i,cl+cr,cr)+dp(i,r,cl,cr+cl)+1ll*(cl+cr)*a[i]);
	return ans;
}
int main()
{
	n=read();
	for(int i=1;i<=n;i++) a[i]=read();
	printf("%lld\n",dp(1,n,1,1)+a[1]+a[n]);
}
```