#include <bits/stdc++.h>

using namespace std;

int seq[19];
int N;

long long dfs(int l,int r,long long lt,long long rt)
{
	if(l + 1 == r) return 0;
	long long ans = 1e18;
	long long tms = lt + rt;
	for(int i = l + 1;i < r; ++ i) ans = min(ans,dfs(l,i,lt,tms) + dfs(i,r,tms,rt) + tms * seq[i]);
	return ans;
}

int main()
{
	scanf("%d",&N);
	for(int i = 1;i <= N; ++ i) scanf("%d",&seq[i]);
	printf("%lld\n",dfs(1,N,1,1) + seq[1] + seq[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 <bits/stdc++.h>

using namespace std;

int seq[19];
int N;

long long dfs(int l,int r,long long lt,long long rt)
{
	if(l + 1 == r) return 0;
	long long ans = 1e18;
	long long tms = lt + rt;
	for(int i = l + 1;i < r; ++ i) ans = min(ans,dfs(l,i,lt,tms) + dfs(i,r,tms,rt) + tms * seq[i]);
	return ans;
}

int main()
{
	scanf("%d",&N);
	for(int i = 1;i <= N; ++ i) scanf("%d",&seq[i]);
	printf("%lld\n",dfs(1,N,1,1) + seq[1] + seq[N]);
}
```