#include<bits/stdc++.h>
#include <list>
using namespace std;
using ll=long long;
const ll MAX=3e2+5,mod=998244353,inf=0x3f3f3f3f3f3f3f3f;
#define X first
#define Y second
#define pb push_back

typedef pair < int , int > ii;
int n;
ll a[30];
ll solve(int l,int r,int p1,int p2){
    if(l+1==r) return 0;
    ll temp=inf;
    for(int i=l+1;i<r;i++) {
        temp=min(temp,solve(l,i,p1,p1+p2)+solve(i,r,p1+p2,p2)+a[i]*(p1+p2));
    }
    return temp;
}
int main(){
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //freopen("agc046.inp","r",stdin);
    //freopen("agc046.out","w",stdout);
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    cout<<solve(1,n,1,1)+a[1]+a[n];
}

### Prompt

Create a solution in Cpp for 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>
#include <list>
using namespace std;
using ll=long long;
const ll MAX=3e2+5,mod=998244353,inf=0x3f3f3f3f3f3f3f3f;
#define X first
#define Y second
#define pb push_back

typedef pair < int , int > ii;
int n;
ll a[30];
ll solve(int l,int r,int p1,int p2){
    if(l+1==r) return 0;
    ll temp=inf;
    for(int i=l+1;i<r;i++) {
        temp=min(temp,solve(l,i,p1,p1+p2)+solve(i,r,p1+p2,p2)+a[i]*(p1+p2));
    }
    return temp;
}
int main(){
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //freopen("agc046.inp","r",stdin);
    //freopen("agc046.out","w",stdout);
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    cout<<solve(1,n,1,1)+a[1]+a[n];
}

```