#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long

using namespace std;

int n;
ll a[20];

ll dp(int ,int ,int ,int);

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

ll dp(int l, int r, int x, int y){
	if(r - l + 1 == 2) return a[l] * x + a[r] * y;
	ll ans = 999999999999999;
	for(int i = l + 1; i < r; ++i)
		ans = min(ans, dp(l, i, x, x + y) + dp(i, r, x + y, y) - a[i] * (x + y));
	return ans;
}
### 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 <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long

using namespace std;

int n;
ll a[20];

ll dp(int ,int ,int ,int);

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

ll dp(int l, int r, int x, int y){
	if(r - l + 1 == 2) return a[l] * x + a[r] * y;
	ll ans = 999999999999999;
	for(int i = l + 1; i < r; ++i)
		ans = min(ans, dp(l, i, x, x + y) + dp(i, r, x + y, y) - a[i] * (x + y));
	return ans;
}
```