#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define x first
#define y second
typedef pair<int, int> pii;
typedef long long ll;
const int MOD = 1000000007;
const int N = 18;
int n,a[N];

ll calc(int l , int x , int r , int y){
  if(r-l==1)return 0;
  ll ret = 1e18;
  for(int idx = l+1 ; idx < r ; ++idx){
    ret = min(ret,calc(l,x,idx,(x+y)) + calc(idx,(x+y),r,y) + (x+y+0ll)*a[idx]);
  }
  return ret;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i)scanf("%d",a+i);
  ll ans = 1ll*a[0]+a[n-1]+calc(0,1,n-1,1);
  cout << ans << endl;
  return 0;
}
### Prompt

Your task is to create 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;
#define pb push_back
#define mp make_pair
#define x first
#define y second
typedef pair<int, int> pii;
typedef long long ll;
const int MOD = 1000000007;
const int N = 18;
int n,a[N];

ll calc(int l , int x , int r , int y){
  if(r-l==1)return 0;
  ll ret = 1e18;
  for(int idx = l+1 ; idx < r ; ++idx){
    ret = min(ret,calc(l,x,idx,(x+y)) + calc(idx,(x+y),r,y) + (x+y+0ll)*a[idx]);
  }
  return ret;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i)scanf("%d",a+i);
  ll ans = 1ll*a[0]+a[n-1]+calc(0,1,n-1,1);
  cout << ans << endl;
  return 0;
}
```