#include<bits/stdc++.h>
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;

typedef long long LL;
typedef pair<pair<int,int>,pair<LL,LL>> pr;

const int maxn=20;

int n;
LL a[maxn];

map<pr,LL> f;
LL dfs(int l,int r,LL x,LL y)
{
	if (l+1>=r) return 0;
	pr now=make_pair(make_pair(l,r),make_pair(x,y));
	if (f.count(now)) return f[now];
	LL re=5e18;
	fo(i,l+1,r-1) re=min(re,dfs(l,i,x,x+y)+dfs(i,r,x+y,y)+a[i]*(x+y));
	return f[now]=re;
}

int main()
{
	scanf("%d",&n);
	fo(i,1,n) scanf("%lld",&a[i]);
	
	printf("%lld\n",dfs(1,n,1,1)+a[1]+a[n]);
}

### Prompt

Your challenge is to write 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
#include<bits/stdc++.h>
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;

typedef long long LL;
typedef pair<pair<int,int>,pair<LL,LL>> pr;

const int maxn=20;

int n;
LL a[maxn];

map<pr,LL> f;
LL dfs(int l,int r,LL x,LL y)
{
	if (l+1>=r) return 0;
	pr now=make_pair(make_pair(l,r),make_pair(x,y));
	if (f.count(now)) return f[now];
	LL re=5e18;
	fo(i,l+1,r-1) re=min(re,dfs(l,i,x,x+y)+dfs(i,r,x+y,y)+a[i]*(x+y));
	return f[now]=re;
}

int main()
{
	scanf("%d",&n);
	fo(i,1,n) scanf("%lld",&a[i]);
	
	printf("%lld\n",dfs(1,n,1,1)+a[1]+a[n]);
}

```