#include <algorithm>
#include <cstring>
#include <cstdio>
#define int long long

const int N = 22;

int n, a[N];
int dfs(int l, int r, int x, int y) {
	int ans = 1e18;
	if (r - l <= 1) return 0;
	for (int k = l + 1; k <= r - 1; k++)
		ans = std::min(ans, dfs(l, k, x, x + y) + dfs(k, r, x + y, y) + (x + y) * a[k]);
	return ans;
}
signed main() {
	scanf("%lld", &n);
	for (int i = 1; i <= n; i++) scanf("%lld", a + i);
	printf("%lld", a[1] + a[n] + dfs(1, n, 1, 1));
}
### Prompt

Please formulate 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 <algorithm>
#include <cstring>
#include <cstdio>
#define int long long

const int N = 22;

int n, a[N];
int dfs(int l, int r, int x, int y) {
	int ans = 1e18;
	if (r - l <= 1) return 0;
	for (int k = l + 1; k <= r - 1; k++)
		ans = std::min(ans, dfs(l, k, x, x + y) + dfs(k, r, x + y, y) + (x + y) * a[k]);
	return ans;
}
signed main() {
	scanf("%lld", &n);
	for (int i = 1; i <= n; i++) scanf("%lld", a + i);
	printf("%lld", a[1] + a[n] + dfs(1, n, 1, 1));
}
```