//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
using namespace std;
typedef long long ll;
const int N=25;
int n,a[N];
ll dfs(int l,int r,ll xl,ll xr){
	if(l+1==r)return 0;
	R ll res=1e18;
	fp(m,l+1,r-1)cmin(res,dfs(l,m,xl,xl+xr)+dfs(m,r,xl+xr,xr)+(xl+xr)*a[m]);
	return res;
}
int main(){
	scanf("%d",&n);
	fp(i,1,n)scanf("%d",&a[i]);
	printf("%lld\n",dfs(1,n,1,1)+a[1]+a[n]);
	return 0;
}
### 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
//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
using namespace std;
typedef long long ll;
const int N=25;
int n,a[N];
ll dfs(int l,int r,ll xl,ll xr){
	if(l+1==r)return 0;
	R ll res=1e18;
	fp(m,l+1,r-1)cmin(res,dfs(l,m,xl,xl+xr)+dfs(m,r,xl+xr,xr)+(xl+xr)*a[m]);
	return res;
}
int main(){
	scanf("%d",&n);
	fp(i,1,n)scanf("%d",&a[i]);
	printf("%lld\n",dfs(1,n,1,1)+a[1]+a[n]);
	return 0;
}
```