#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int const N = 18;
map<pair<int, int>, ll> dp[17][17];
int x[N];

ll go(int l, int r, int cl, int cr){
	if (l > r)return 0;
	if (l == r)return (cl + cr) * (ll)x[l];
	auto it = dp[l][r].find(make_pair(cl, cr));
	if (it != dp[l][r].end())return it->second;
	ll mn = 1e18;
	f(i, l, r + 1)mn = min(mn, go(l, i - 1, cl, cl + cr) + go(i + 1, r, cl + cr, cr) + (cl + cr) * (ll)x[i]);
	return dp[l][r][make_pair(cl, cr)] = mn;
}

int main(){
	int n;
	scanf("%d", &n);
	f(i, 0, n)scanf("%d", x + i);
	printf("%lld\n", go(1, n - 2, 1, 1) + x[0] + x[n - 1]);
}
### Prompt

Develop a solution in Cpp 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>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int const N = 18;
map<pair<int, int>, ll> dp[17][17];
int x[N];

ll go(int l, int r, int cl, int cr){
	if (l > r)return 0;
	if (l == r)return (cl + cr) * (ll)x[l];
	auto it = dp[l][r].find(make_pair(cl, cr));
	if (it != dp[l][r].end())return it->second;
	ll mn = 1e18;
	f(i, l, r + 1)mn = min(mn, go(l, i - 1, cl, cl + cr) + go(i + 1, r, cl + cr, cr) + (cl + cr) * (ll)x[i]);
	return dp[l][r][make_pair(cl, cr)] = mn;
}

int main(){
	int n;
	scanf("%d", &n);
	f(i, 0, n)scanf("%d", x + i);
	printf("%lld\n", go(1, n - 2, 1, 1) + x[0] + x[n - 1]);
}
```