#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
const int MAXN = 20;
const ll INF = (1LL<<60);
int N; ll A[MAXN + 5];
ll dfs(int l, int r, int cntl, int cntr) {
	if( l + 1 == r ) return 0;
	ll ret = dfs(l, l+1, cntl, cntl+cntr) + dfs(l+1, r, cntl+cntr, cntr) + (cntl+cntr)*A[l+1];
	for(int i=l+2;i<=r-1;i++)
		ret = min(ret, dfs(l, i, cntl, cntl+cntr) + dfs(i, r, cntl+cntr, cntr) + (cntl+cntr)*A[i]);
	return ret;
}
int main() {
	scanf("%d", &N);
	for(int i=1;i<=N;i++)
		scanf("%lld", &A[i]);
	printf("%lld\n", dfs(1, N, 1, 1) + A[1] + A[N]);
}
### 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>
using namespace std;
typedef long long ll;
const int MAXN = 20;
const ll INF = (1LL<<60);
int N; ll A[MAXN + 5];
ll dfs(int l, int r, int cntl, int cntr) {
	if( l + 1 == r ) return 0;
	ll ret = dfs(l, l+1, cntl, cntl+cntr) + dfs(l+1, r, cntl+cntr, cntr) + (cntl+cntr)*A[l+1];
	for(int i=l+2;i<=r-1;i++)
		ret = min(ret, dfs(l, i, cntl, cntl+cntr) + dfs(i, r, cntl+cntr, cntr) + (cntl+cntr)*A[i]);
	return ret;
}
int main() {
	scanf("%d", &N);
	for(int i=1;i<=N;i++)
		scanf("%lld", &A[i]);
	printf("%lld\n", dfs(1, N, 1, 1) + A[1] + A[N]);
}
```