#include <cstdio>
#include <algorithm>
using namespace std;

typedef long long ll;
const int Maxn = 18;
const ll INF = 0x3f3f3f3f3f3f3f3f;

int N, A[Maxn + 5];

ll DFS(int lef, int rig, int cl, int cr) {
	if(rig - lef + 1 <= 2) return 0;
	ll ret = INF;
	for(int i = lef + 1; i < rig; i++)
		ret = min(ret, DFS(lef, i, cl, cl + cr) + DFS(i, rig, cl + cr, cr) + 1LL * (cl + cr) * A[i]);
	return ret;
}

int main() {
#ifdef LOACL
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif
	scanf("%d", &N);
	for(int i = 1; i <= N; i++)
		scanf("%d", &A[i]);
	printf("%lld\n", DFS(1, N, 1, 1) + A[1] + A[N]);
	return 0;
}
### Prompt

Please create a solution in Cpp 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 <algorithm>
using namespace std;

typedef long long ll;
const int Maxn = 18;
const ll INF = 0x3f3f3f3f3f3f3f3f;

int N, A[Maxn + 5];

ll DFS(int lef, int rig, int cl, int cr) {
	if(rig - lef + 1 <= 2) return 0;
	ll ret = INF;
	for(int i = lef + 1; i < rig; i++)
		ret = min(ret, DFS(lef, i, cl, cl + cr) + DFS(i, rig, cl + cr, cr) + 1LL * (cl + cr) * A[i]);
	return ret;
}

int main() {
#ifdef LOACL
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif
	scanf("%d", &N);
	for(int i = 1; i <= N; i++)
		scanf("%d", &A[i]);
	printf("%lld\n", DFS(1, N, 1, 1) + A[1] + A[N]);
	return 0;
}
```