#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
int n;
ll a[ 19 ];

ll calc(int l, int r, ll xl, ll xr) {
      if (r - l == 1) return a[l] * xl + a[r] * xr;
      ll ans = 1ll << 60;
      for (int i = l + 1; i < r; i++) {
        ans = min(ans, calc(l, i, xl, xl + xr) + calc(i, r, xl + xr, xr) - (xl + xr) * a[i]);
      }
      return ans;
    }


int main()
{
    scanf( "%d", &n );
    for( int i = 1; i <= n; i ++ )
    {
        scanf( "%lld", &a[ i ] );
    }
    printf( "%lld", calc( 1, n, 1, 1 ) );
}







### 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;
typedef long long ll;
int n;
ll a[ 19 ];

ll calc(int l, int r, ll xl, ll xr) {
      if (r - l == 1) return a[l] * xl + a[r] * xr;
      ll ans = 1ll << 60;
      for (int i = l + 1; i < r; i++) {
        ans = min(ans, calc(l, i, xl, xl + xr) + calc(i, r, xl + xr, xr) - (xl + xr) * a[i]);
      }
      return ans;
    }


int main()
{
    scanf( "%d", &n );
    for( int i = 1; i <= n; i ++ )
    {
        scanf( "%lld", &a[ i ] );
    }
    printf( "%lld", calc( 1, n, 1, 1 ) );
}







```