#include <bits/stdc++.h>

typedef long long ll;

int n;
ll a[20];

inline void down(ll &x, const ll y) {x > y ? x = y : 0;}

ll dfs(int l, int r, int cl, int cr) {
	if (l > r) return 0;
	int i; ll ret = LLONG_MAX;
	for (i = l; i <= r; ++i) down(ret, (cl + cr) * a[i] + dfs(l, i - 1, cl, cl + cr) + dfs(i + 1, r, cl + cr, cr));
	return ret;
}

int main() {
	int i;
	scanf("%d", &n);
	for (i = 1; i <= n; ++i) scanf("%lld", a + i);
	printf("%lld\n", a[1] + a[n] + dfs(2, n - 1, 1, 1));
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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>

typedef long long ll;

int n;
ll a[20];

inline void down(ll &x, const ll y) {x > y ? x = y : 0;}

ll dfs(int l, int r, int cl, int cr) {
	if (l > r) return 0;
	int i; ll ret = LLONG_MAX;
	for (i = l; i <= r; ++i) down(ret, (cl + cr) * a[i] + dfs(l, i - 1, cl, cl + cr) + dfs(i + 1, r, cl + cr, cr));
	return ret;
}

int main() {
	int i;
	scanf("%d", &n);
	for (i = 1; i <= n; ++i) scanf("%lld", a + i);
	printf("%lld\n", a[1] + a[n] + dfs(2, n - 1, 1, 1));
	return 0;
}
```