#include <cstdio>
#include <algorithm>
#include <map>

typedef long long LL;
const int MN = 20;

int N, A[MN];

LL F(int l, int r, int cl, int cr) {
	if (l > r) return 0;
	LL ans = 0x3f3f3f3f3f3f3f3f;
	for (int i = l; i <= r; ++i)
		ans = std::min(ans, F(l, i - 1, cl, cl + cr) + (LL)A[i] * (cl + cr) + F(i + 1, r, cl + cr, cr));
	return ans;
}

int main() {
	scanf("%d", &N);
	for (int i = 0; i < N; ++i) scanf("%d", &A[i]);
	N -= 2;
	printf("%lld\n", F(1, N, 1, 1) + A[0] + A[N + 1]);
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 <algorithm>
#include <map>

typedef long long LL;
const int MN = 20;

int N, A[MN];

LL F(int l, int r, int cl, int cr) {
	if (l > r) return 0;
	LL ans = 0x3f3f3f3f3f3f3f3f;
	for (int i = l; i <= r; ++i)
		ans = std::min(ans, F(l, i - 1, cl, cl + cr) + (LL)A[i] * (cl + cr) + F(i + 1, r, cl + cr, cr));
	return ans;
}

int main() {
	scanf("%d", &N);
	for (int i = 0; i < N; ++i) scanf("%d", &A[i]);
	N -= 2;
	printf("%lld\n", F(1, N, 1, 1) + A[0] + A[N + 1]);
	return 0;
}
```