#include <bits/stdc++.h>

using namespace std; 

int n, a[20]; 

map < pair <pair <int, int>, pair <long long, long long> > , long long> f, vis; 

long long dp(int l, int r, long long x, long long y) {
  if (vis[{{l, r}, {x, y}}] != 0) return f[{{l, r}, {x, y}}];
  long long ret = 1e18; 
  vis[{{l, r}, {x, y}}] = 1; 
  if (r == l + 1) {
    ret = a[l] * x + a[r] * y; 
  }
  else {
    for (int m = l + 1; m <= r - 1; m++) {
      ret = min(ret, dp(l, m, x, x + y) + dp(m, r, x + y, y) - 1ll * a[m] * (x + y)); 
    }
  }
  return f[{{l, r}, {x, y}}] = ret; 
}

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

Your challenge is to write 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; 

int n, a[20]; 

map < pair <pair <int, int>, pair <long long, long long> > , long long> f, vis; 

long long dp(int l, int r, long long x, long long y) {
  if (vis[{{l, r}, {x, y}}] != 0) return f[{{l, r}, {x, y}}];
  long long ret = 1e18; 
  vis[{{l, r}, {x, y}}] = 1; 
  if (r == l + 1) {
    ret = a[l] * x + a[r] * y; 
  }
  else {
    for (int m = l + 1; m <= r - 1; m++) {
      ret = min(ret, dp(l, m, x, x + y) + dp(m, r, x + y, y) - 1ll * a[m] * (x + y)); 
    }
  }
  return f[{{l, r}, {x, y}}] = ret; 
}

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