#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
using namespace std;
#define N 200012
typedef long long ll;
struct node{ll x,y;friend bool operator<(node a,node b){return (a.x^b.x)?a.x<b.x:a.y<b.y;}};
int n,a[N];
map<node,ll>dp[22][22];
ll cal(int l,int r,ll x,ll y)
{if(l==r-1)return 0;if(dp[l][r].count((node){x,y}))return dp[l][r][(node){x,y}];
	ll ret=4e18;
	for(int i=l+1;i<r;i++)
	{
		ret=min(ret,cal(l,i,x,x+y)+cal(i,r,x+y,y)+1ll*(x+y)*a[i]);
	}return dp[l][r][(node){x,y}]=ret;
}
int main()
{
    scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	printf("%lld\n",cal(1,n,1,1)+a[1]+a[n]);
}

### Prompt

Please create a solution in Cpp 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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
using namespace std;
#define N 200012
typedef long long ll;
struct node{ll x,y;friend bool operator<(node a,node b){return (a.x^b.x)?a.x<b.x:a.y<b.y;}};
int n,a[N];
map<node,ll>dp[22][22];
ll cal(int l,int r,ll x,ll y)
{if(l==r-1)return 0;if(dp[l][r].count((node){x,y}))return dp[l][r][(node){x,y}];
	ll ret=4e18;
	for(int i=l+1;i<r;i++)
	{
		ret=min(ret,cal(l,i,x,x+y)+cal(i,r,x+y,y)+1ll*(x+y)*a[i]);
	}return dp[l][r][(node){x,y}]=ret;
}
int main()
{
    scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	printf("%lld\n",cal(1,n,1,1)+a[1]+a[n]);
}

```