#include <bits/stdc++.h>
using namespace std;
#define finish(x) return cout << x << endl, 0
#define ll long long

int n;
vector <int> a;

ll solve(int l, ll cl, int r, ll cr){
    if(l + 1 == r) return 0;
    ll ret = 1e18;
    for(int k = l + 1 ; k < r ; k++){
        ret = min(ret, solve(l, cl, k, cl + cr) + solve(k, cl + cr, r, cr) + (cl + cr) * a[k]);
    }
    return ret;
}
int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    a.resize(n);
    for(auto &i : a) cin >> i;
    cout << solve(0, 1, n - 1, 1) + a[0] + a[n - 1] << endl;
}

### 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;
#define finish(x) return cout << x << endl, 0
#define ll long long

int n;
vector <int> a;

ll solve(int l, ll cl, int r, ll cr){
    if(l + 1 == r) return 0;
    ll ret = 1e18;
    for(int k = l + 1 ; k < r ; k++){
        ret = min(ret, solve(l, cl, k, cl + cr) + solve(k, cl + cr, r, cr) + (cl + cr) * a[k]);
    }
    return ret;
}
int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    a.resize(n);
    for(auto &i : a) cin >> i;
    cout << solve(0, 1, n - 1, 1) + a[0] + a[n - 1] << endl;
}

```