#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n;
int a[20];
map<int,ll> dp;
ll solve(int l, int r, int vl, int vr)
{
	if (l>r) return 0;
	ll hs = 0;
	hs = hs*23+l;
	hs = hs*23+r;
	hs = hs*23333333+vl;
	hs = hs*23333333+vr;
	if (dp.find(hs)!=dp.end()) return dp[hs];
	ll &cur = dp[hs];
	cur = 8e18;
	for (int i=l; i<=r; i++)
	{
		ll val = solve(l, i-1, vl, vl+vr)+solve(i+1, r, vl+vr, vr)+1ll*a[i]*(vl+vr);
		cur = min(cur, val);
	}
	return cur;
}
int main()
{
	cin>>n;
	for (int i=1; i<=n; i++) cin>>a[i];
	cout<<solve(2, n-1, 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>
using namespace std;
typedef long long ll;
int n;
int a[20];
map<int,ll> dp;
ll solve(int l, int r, int vl, int vr)
{
	if (l>r) return 0;
	ll hs = 0;
	hs = hs*23+l;
	hs = hs*23+r;
	hs = hs*23333333+vl;
	hs = hs*23333333+vr;
	if (dp.find(hs)!=dp.end()) return dp[hs];
	ll &cur = dp[hs];
	cur = 8e18;
	for (int i=l; i<=r; i++)
	{
		ll val = solve(l, i-1, vl, vl+vr)+solve(i+1, r, vl+vr, vr)+1ll*a[i]*(vl+vr);
		cur = min(cur, val);
	}
	return cur;
}
int main()
{
	cin>>n;
	for (int i=1; i<=n; i++) cin>>a[i];
	cout<<solve(2, n-1, 1, 1)+a[1]+a[n]<<endl;
	return 0;
}
```