#include <bits/stdc++.h>
#define N 20

using namespace std;

int n;
long long a[N];

long long f(int l, int r, long long x, long long y) {
	if(l + 1 == r) return 0;
	long long res = 1e18;
	for(int i = l + 1; i <= r - 1; ++i) {
		res = min(res, 
		f(l, i, x, x + y) + 
		f(i, r, x + y, y) + 
		(x + y) * a[i]);
	}
	return res;
}

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

Your challenge is to write a CPP solution 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 <bits/stdc++.h>
#define N 20

using namespace std;

int n;
long long a[N];

long long f(int l, int r, long long x, long long y) {
	if(l + 1 == r) return 0;
	long long res = 1e18;
	for(int i = l + 1; i <= r - 1; ++i) {
		res = min(res, 
		f(l, i, x, x + y) + 
		f(i, r, x + y, y) + 
		(x + y) * a[i]);
	}
	return res;
}

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