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

using PP = pair<int, int>;
int n;
long a[18];
map<PP, long> memo[18][18];
long rec(int l, int r, int p, int q) {
  if (l + 1 == r) return p * a[l] + q * a[r];
  const auto it = memo[l][r].find(PP(p, q));
  if (it != memo[l][r].end()) return it->second;
  
  long mn = 1e18;
  for (int m = l + 1; m < r; ++m) {
    long nv = rec(l, m, p, p + q) + rec(m, r, p + q, q) - a[m] * (p + q);
    mn = min(mn, nv);
  }
  return mn;
}

int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  cout << rec(0, n - 1, 1, 1) << endl;
}
### Prompt

Generate 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;

using PP = pair<int, int>;
int n;
long a[18];
map<PP, long> memo[18][18];
long rec(int l, int r, int p, int q) {
  if (l + 1 == r) return p * a[l] + q * a[r];
  const auto it = memo[l][r].find(PP(p, q));
  if (it != memo[l][r].end()) return it->second;
  
  long mn = 1e18;
  for (int m = l + 1; m < r; ++m) {
    long nv = rec(l, m, p, p + q) + rec(m, r, p + q, q) - a[m] * (p + q);
    mn = min(mn, nv);
  }
  return mn;
}

int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  cout << rec(0, n - 1, 1, 1) << endl;
}
```