#include <bits/stdc++.h>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define dow(i,j,k) for (int i = j; i >= k; i--)
#define fi first
#define se second
#define mp make_pair
#define pb push_back
using namespace std;
typedef pair<int, int> pi;
typedef long long ll;
int n;
ll a[20];
long long ans;
ll find(int l, int xl, int r, int xr) {
	if (l + 1 == r) return 0;
	long long res = 1e18;
	for (int i = l + 1; i < r; i++) 
		res = min(res, find(l, xl, i, xl + xr) + find(i, xl + xr, r, xr) + a[i] * (xl + xr));
	return res;
}
		
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) cin >> a[i];
	cout << find(1, 1, n, 1) + a[1] + a[n] << endl;
}
### Prompt

Create a solution in CPP for 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>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define dow(i,j,k) for (int i = j; i >= k; i--)
#define fi first
#define se second
#define mp make_pair
#define pb push_back
using namespace std;
typedef pair<int, int> pi;
typedef long long ll;
int n;
ll a[20];
long long ans;
ll find(int l, int xl, int r, int xr) {
	if (l + 1 == r) return 0;
	long long res = 1e18;
	for (int i = l + 1; i < r; i++) 
		res = min(res, find(l, xl, i, xl + xr) + find(i, xl + xr, r, xr) + a[i] * (xl + xr));
	return res;
}
		
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) cin >> a[i];
	cout << find(1, 1, n, 1) + a[1] + a[n] << endl;
}
```