#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<iomanip>
#include<cstring>
#include<algorithm>
#include<ctime>
using namespace std;
#define ll long long
int n, A[25];
ll Getans(int l, int r, ll L, ll R)
{
	if (l + 1 == r)
		return L * A[l] + R * A[r];
	ll ret = 0x7f7f7f7f7f7f7f7fll;
	for (int i = l + 1; i < r; i++)
		ret = min(Getans(l, i, L, L + R) + Getans(i, r, L + R, R) - A[i] * (L + R), ret);
	return ret;
}
int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d", &A[i]);
	printf("%lld\n", Getans(1, n, 1, 1));
}

### 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<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<iomanip>
#include<cstring>
#include<algorithm>
#include<ctime>
using namespace std;
#define ll long long
int n, A[25];
ll Getans(int l, int r, ll L, ll R)
{
	if (l + 1 == r)
		return L * A[l] + R * A[r];
	ll ret = 0x7f7f7f7f7f7f7f7fll;
	for (int i = l + 1; i < r; i++)
		ret = min(Getans(l, i, L, L + R) + Getans(i, r, L + R, R) - A[i] * (L + R), ret);
	return ret;
}
int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d", &A[i]);
	printf("%lld\n", Getans(1, n, 1, 1));
}

```