#include <bits/stdc++.h>
#define ll long long
#define INF 1e15

using namespace std;

int n, a[20];
int id[2010][2010], tot;
ll dp[20][20][1 << 18];

ll solve(int l, int r, int x, int y)
{
	if (l + 1 == r) return 0;
	if (!id[x][y]) id[x][y] = ++tot;
	if (dp[l][r][id[x][y]] != -1) return dp[l][r][id[x][y]];
	ll ret = INF;
	for (int i = l + 1; i < r; i++)
		ret = min(ret, solve(l, i, x, x + y) + solve(i, r, x + y, y) + 1ll * a[i] * (x + y));
	return dp[l][r][id[x][y]] = ret;
}

int main()
{
	ios::sync_with_stdio(false);
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	memset(dp, -1, sizeof dp);
	cout << solve(1, n, 1, 1) + a[1] + a[n] << 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
#include <bits/stdc++.h>
#define ll long long
#define INF 1e15

using namespace std;

int n, a[20];
int id[2010][2010], tot;
ll dp[20][20][1 << 18];

ll solve(int l, int r, int x, int y)
{
	if (l + 1 == r) return 0;
	if (!id[x][y]) id[x][y] = ++tot;
	if (dp[l][r][id[x][y]] != -1) return dp[l][r][id[x][y]];
	ll ret = INF;
	for (int i = l + 1; i < r; i++)
		ret = min(ret, solve(l, i, x, x + y) + solve(i, r, x + y, y) + 1ll * a[i] * (x + y));
	return dp[l][r][id[x][y]] = ret;
}

int main()
{
	ios::sync_with_stdio(false);
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	memset(dp, -1, sizeof dp);
	cout << solve(1, n, 1, 1) + a[1] + a[n] << endl;
	return 0;
}

```