#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int,int>

int n, org[22];
map<pii,ll>mp[22][22];

ll gao(int l,int r,int a,int b){
    if (l+1==r) return 0;
    if (l+2==r) return org[l+1]*(ll)(a+b);
    if (mp[l][r].count({a,b})) return mp[l][r][{a,b}];
    ll res=1e100;
    for (int i=l+1;i<=r-1;++i){
        ll t=gao(l,i,a,a+b)+org[i]*(ll)(a+b)+gao(i,r,a+b,b);
        res=min(res,t);
    }
    return mp[l][r][{a,b}]=res;
}

int main(){
    cin>>n;
    for (int i=1;i<=n;++i)
        cin>>org[i];
    cout<<gao(1,n,1,1)+org[1]+org[n]<<endl;
}
### Prompt

Your task is to create 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 ll long long
#define pii pair<int,int>

int n, org[22];
map<pii,ll>mp[22][22];

ll gao(int l,int r,int a,int b){
    if (l+1==r) return 0;
    if (l+2==r) return org[l+1]*(ll)(a+b);
    if (mp[l][r].count({a,b})) return mp[l][r][{a,b}];
    ll res=1e100;
    for (int i=l+1;i<=r-1;++i){
        ll t=gao(l,i,a,a+b)+org[i]*(ll)(a+b)+gao(i,r,a+b,b);
        res=min(res,t);
    }
    return mp[l][r][{a,b}]=res;
}

int main(){
    cin>>n;
    for (int i=1;i<=n;++i)
        cin>>org[i];
    cout<<gao(1,n,1,1)+org[1]+org[n]<<endl;
}
```