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

typedef long long ll;
const int N=20;
int a[N],n;
ll dfs(int L,int R,int lt,int rt){
	if(L>R)return 0;
	if(L==R)return (ll)(lt+rt)*a[L];
	ll ret=LLONG_MAX;
	for(int i=L;i<=R;++i)
		ret=min(ret,dfs(L,i-1,lt,lt+rt)+dfs(i+1,R,lt+rt,rt)+(ll)(lt+rt)*a[i]);
	return ret;
}

int main(){
#ifdef ONLINE_JUDGE
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#endif
	cin>>n;
	for(int i=1;i<=n;++i)cin>>a[i];
	cout<<dfs(2,n-1,1,1)+a[1]+a[n]<<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;
const int N=20;
int a[N],n;
ll dfs(int L,int R,int lt,int rt){
	if(L>R)return 0;
	if(L==R)return (ll)(lt+rt)*a[L];
	ll ret=LLONG_MAX;
	for(int i=L;i<=R;++i)
		ret=min(ret,dfs(L,i-1,lt,lt+rt)+dfs(i+1,R,lt+rt,rt)+(ll)(lt+rt)*a[i]);
	return ret;
}

int main(){
#ifdef ONLINE_JUDGE
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#endif
	cin>>n;
	for(int i=1;i<=n;++i)cin>>a[i];
	cout<<dfs(2,n-1,1,1)+a[1]+a[n]<<endl;
	return 0;
}


```