// D - Add and Remove
#include <bits/stdc++.h>

using namespace std;

long long dfs(const vector<long long> &a, int l, int r, long long lc, long long rc)
{
	if (r - l <= 1) return 0;
	long long best = 1e18;
	for (int m = l + 1; m < r; m++) {
		long long s = a[m] * (lc + rc) + dfs(a, l, m, lc, lc + rc) + dfs(a, m, r, lc + rc, rc);
		best = min(best, s);
	}
	return best;
}

int main(int argc, char *argv[])
{
	ios::sync_with_stdio(false); cin.tie(0);
	int n;
	cin >> n;
	vector<long long> a(n);
	for (int i = 0; i < n; i++) cin >> a[i];
	cout << dfs(a, 0, n - 1, 1, 1) + a[0] + a[n - 1] << endl;
	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
// D - Add and Remove
#include <bits/stdc++.h>

using namespace std;

long long dfs(const vector<long long> &a, int l, int r, long long lc, long long rc)
{
	if (r - l <= 1) return 0;
	long long best = 1e18;
	for (int m = l + 1; m < r; m++) {
		long long s = a[m] * (lc + rc) + dfs(a, l, m, lc, lc + rc) + dfs(a, m, r, lc + rc, rc);
		best = min(best, s);
	}
	return best;
}

int main(int argc, char *argv[])
{
	ios::sync_with_stdio(false); cin.tie(0);
	int n;
	cin >> n;
	vector<long long> a(n);
	for (int i = 0; i < n; i++) cin >> a[i];
	cout << dfs(a, 0, n - 1, 1, 1) + a[0] + a[n - 1] << endl;
	return 0;
}

```