#include <bits/stdc++.h>

typedef long long LL;
const int MAXN = 18;
int A[MAXN], n;
LL dp(int l, int r, LL vl, LL vr) {
	if (l > r) return 0;
	if (l == r) return (vl + vr) * A[l];
	LL res = 0x7fffffffffffffffLL;
	for (int i = l; i <= r; ++i)
		res = std::min(res, dp(l, i - 1, vl, vl + vr) + dp(i + 1, r, vl + vr, vr) + (vl + vr) * A[i]);
	return res;
}
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1; i <= n; ++i) std::cin >> A[i];
	std::cout << dp(2, n - 1, 1, 1) + A[1] + A[n] << std::endl;
	return 0;
}

### 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>

typedef long long LL;
const int MAXN = 18;
int A[MAXN], n;
LL dp(int l, int r, LL vl, LL vr) {
	if (l > r) return 0;
	if (l == r) return (vl + vr) * A[l];
	LL res = 0x7fffffffffffffffLL;
	for (int i = l; i <= r; ++i)
		res = std::min(res, dp(l, i - 1, vl, vl + vr) + dp(i + 1, r, vl + vr, vr) + (vl + vr) * A[i]);
	return res;
}
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1; i <= n; ++i) std::cin >> A[i];
	std::cout << dp(2, n - 1, 1, 1) + A[1] + A[n] << std::endl;
	return 0;
}

```