#include <bits/stdc++.h>

using namespace std;

int const N = 20;
long long const MAX = 1e18 + 1;

int n;
long long a[N];

long long calc(int l, int r, long long costL, long long costR) {
	if (r - l <= 1) {
		return 0;
	}
	long long res = MAX;
	for (int mid = l + 1; mid < r; mid++) {
		res = min(res, a[mid] * (costL + costR) + calc(l, mid, costL, costR + costL) + calc(mid, r, costL + costR, costR));
	}
	return res;
}

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

In Cpp, your task is to solve 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 const N = 20;
long long const MAX = 1e18 + 1;

int n;
long long a[N];

long long calc(int l, int r, long long costL, long long costR) {
	if (r - l <= 1) {
		return 0;
	}
	long long res = MAX;
	for (int mid = l + 1; mid < r; mid++) {
		res = min(res, a[mid] * (costL + costR) + calc(l, mid, costL, costR + costL) + calc(mid, r, costL + costR, costR));
	}
	return res;
}

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