#include <stdio.h>
int n;
long long a[18];
long long f[18];
long long p, q;
void so(int u, int v) {
	if (u == n) {
		p = 0;
		for (int i = 0; i < u; i++) {
			p += a[i] * f[i];
		}
		if (q > p)q = p;
		return;
	}
	for (int i = v; i < u; i++) {
		for (int j = u - 1; j >= i; j--) {
			a[j + 1] = a[j];
		}
		a[i] = a[i - 1] + a[i + 1];
		so(u + 1, i);
		for (int j = i; j < u; j++) {
			a[j] = a[j + 1];
		}
	}
}
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%lld", f + i);
	}
	a[0] = 1;
	a[1] = 1;
	q = 400000000000;
	so(2, 1);
	printf("%lld\n", q);
}
### Prompt

Please formulate 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 <stdio.h>
int n;
long long a[18];
long long f[18];
long long p, q;
void so(int u, int v) {
	if (u == n) {
		p = 0;
		for (int i = 0; i < u; i++) {
			p += a[i] * f[i];
		}
		if (q > p)q = p;
		return;
	}
	for (int i = v; i < u; i++) {
		for (int j = u - 1; j >= i; j--) {
			a[j + 1] = a[j];
		}
		a[i] = a[i - 1] + a[i + 1];
		so(u + 1, i);
		for (int j = i; j < u; j++) {
			a[j] = a[j + 1];
		}
	}
}
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%lld", f + i);
	}
	a[0] = 1;
	a[1] = 1;
	q = 400000000000;
	so(2, 1);
	printf("%lld\n", q);
}
```