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

LL n, a[20];
unordered_map<LL,LL> dp; 

LL dfs(int l,int r,int xl,int xr){
	if(l+1==r) return 0ll;
	LL k = ((LL)xl<<30)+((LL)xr<<10)+(l<<5)+r;
	if(dp[k]) return dp[k];
	dp[k] = 1ll<<55;
	rep(i,l+1,r-1) dp[k] = min(dp[k], dfs(l,i,xl,xl+xr) + dfs(i,r,xl+xr,xr) + a[i]*(xl+xr) );
	return dp[k];
}

int main(){
	cin>>n;
	rep(i,1,n)cin>>a[i];
	cout<<dfs(1,n,1,1)+a[1]+a[n]<<endl;
	return 0;
}
### 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>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
typedef long long LL;

LL n, a[20];
unordered_map<LL,LL> dp; 

LL dfs(int l,int r,int xl,int xr){
	if(l+1==r) return 0ll;
	LL k = ((LL)xl<<30)+((LL)xr<<10)+(l<<5)+r;
	if(dp[k]) return dp[k];
	dp[k] = 1ll<<55;
	rep(i,l+1,r-1) dp[k] = min(dp[k], dfs(l,i,xl,xl+xr) + dfs(i,r,xl+xr,xr) + a[i]*(xl+xr) );
	return dp[k];
}

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