#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

int n,a[20];
map<pair<int,int>,ll>dp[22][22];

ll dfs(int l,int r,ll mul_l,ll mul_r){
	if(l>r)return 0;
	if(dp[l][r].count(make_pair(mul_l,mul_r)))return dp[l][r][make_pair(mul_l,mul_r)];
	ll &ret=dp[l][r][make_pair(mul_l,mul_r)];ret=1e18;
	for(int mid=l;mid<=r;mid++){
		ret=min(ret,dfs(l,mid-1,mul_l,mul_l+mul_r)+dfs(mid+1,r,mul_l+mul_r,mul_r)+a[mid]*(mul_l+mul_r));
	}
	return ret;
}

int main(){
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	cout<<a[1]+a[n]+dfs(2,n-1,1,1)<<endl;
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;

typedef long long ll;

int n,a[20];
map<pair<int,int>,ll>dp[22][22];

ll dfs(int l,int r,ll mul_l,ll mul_r){
	if(l>r)return 0;
	if(dp[l][r].count(make_pair(mul_l,mul_r)))return dp[l][r][make_pair(mul_l,mul_r)];
	ll &ret=dp[l][r][make_pair(mul_l,mul_r)];ret=1e18;
	for(int mid=l;mid<=r;mid++){
		ret=min(ret,dfs(l,mid-1,mul_l,mul_l+mul_r)+dfs(mid+1,r,mul_l+mul_r,mul_r)+a[mid]*(mul_l+mul_r));
	}
	return ret;
}

int main(){
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	cout<<a[1]+a[n]+dfs(2,n-1,1,1)<<endl;
	return 0;
}
```