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

using ll = long long;
vector<ll> a;
ll compute(ll i, ll j, ll x, ll y);

ll dp(ll i, ll j, ll x, ll y){
    return compute(i, j, x, y);
}
ll compute(ll i, ll j, ll x, ll y){
    if(i + 1 == j) return a[i] * x + a[j] * y;
    ll best = 1e18;
    for(int k = i + 1; k < j; k++){
        best = min(best, compute(i, k, x, x+y) + compute(k, j, x+y, y) - a[k] * (x + y));
    }
    return best;
}

int main(){
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    a.resize(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    cout << dp(0, n-1, 1, 1) << '\n';
}

### Prompt

Develop a solution in cpp to the problem described below:

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;

using ll = long long;
vector<ll> a;
ll compute(ll i, ll j, ll x, ll y);

ll dp(ll i, ll j, ll x, ll y){
    return compute(i, j, x, y);
}
ll compute(ll i, ll j, ll x, ll y){
    if(i + 1 == j) return a[i] * x + a[j] * y;
    ll best = 1e18;
    for(int k = i + 1; k < j; k++){
        best = min(best, compute(i, k, x, x+y) + compute(k, j, x+y, y) - a[k] * (x + y));
    }
    return best;
}

int main(){
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    a.resize(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    cout << dp(0, n-1, 1, 1) << '\n';
}

```