#include<bits/stdc++.h>
using namespace std;
inline int read(){
	register int x=0;
	register char ch=getchar();
	while(ch<'0'||ch>'9')	ch=getchar();
	while(ch>='0'&&ch<='9')	x=x*10+(ch^48),ch=getchar();
	return x;
}
typedef pair<int,int>P;
#define ll long long
const int M=1e5+5;
const int N=19;
int n,a[N];
map<P,ll>f[N][N];
ll solve(int l,int r,int x,int y){
	if(f[l][r].count(P(x,y)))	return f[l][r][P(x,y)];
	if(l==r-1)	return 0;
	ll res=LLONG_MAX;
	for(int i=l+1;i<r;i++)
		res=min(res,solve(l,i,x,x+y)+solve(i,r,x+y,y)+1LL*a[i]*(x+y));
	return f[l][r][P(x,y)]=res;
}
int main(){
	n=read();
	for(int i=1;i<=n;i++)	a[i]=read();
	printf("%lld\n",solve(1,n,1,1)+a[1]+a[n]);
	return 0;
}


### Prompt

Create a solution in cpp for 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>
using namespace std;
inline int read(){
	register int x=0;
	register char ch=getchar();
	while(ch<'0'||ch>'9')	ch=getchar();
	while(ch>='0'&&ch<='9')	x=x*10+(ch^48),ch=getchar();
	return x;
}
typedef pair<int,int>P;
#define ll long long
const int M=1e5+5;
const int N=19;
int n,a[N];
map<P,ll>f[N][N];
ll solve(int l,int r,int x,int y){
	if(f[l][r].count(P(x,y)))	return f[l][r][P(x,y)];
	if(l==r-1)	return 0;
	ll res=LLONG_MAX;
	for(int i=l+1;i<r;i++)
		res=min(res,solve(l,i,x,x+y)+solve(i,r,x+y,y)+1LL*a[i]*(x+y));
	return f[l][r][P(x,y)]=res;
}
int main(){
	n=read();
	for(int i=1;i<=n;i++)	a[i]=read();
	printf("%lld\n",solve(1,n,1,1)+a[1]+a[n]);
	return 0;
}


```