#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N=18;
ll a[N];
map<pair<int,int>,ll> dp[N][N];
ll Solve(int l, int r, int mul_l, int mul_r)
{
	if(l>r) return 0;
	if(dp[l][r].count({mul_l,mul_r})) return dp[l][r][{mul_l,mul_r}];
	if(l==r) return a[l]*(mul_l+mul_r);
	ll ans=1e18;
	for(int i=l;i<=r;i++) ans=min(ans,Solve(l,i-1,mul_l,mul_r+mul_l)+Solve(i+1,r,mul_l+mul_r,mul_r)+a[i]*(mul_l+mul_r));
	return ans;
}
int main()
{
	int n;
	scanf("%i",&n);
    for(int i=0;i<n;i++) scanf("%lld",&a[i]);
    printf("%lld\n",Solve(1,n-2,1,1)+a[0]+a[n-1]);
	return 0;
}
### Prompt

In cpp, your task is to solve the following problem:

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
const int N=18;
ll a[N];
map<pair<int,int>,ll> dp[N][N];
ll Solve(int l, int r, int mul_l, int mul_r)
{
	if(l>r) return 0;
	if(dp[l][r].count({mul_l,mul_r})) return dp[l][r][{mul_l,mul_r}];
	if(l==r) return a[l]*(mul_l+mul_r);
	ll ans=1e18;
	for(int i=l;i<=r;i++) ans=min(ans,Solve(l,i-1,mul_l,mul_r+mul_l)+Solve(i+1,r,mul_l+mul_r,mul_r)+a[i]*(mul_l+mul_r));
	return ans;
}
int main()
{
	int n;
	scanf("%i",&n);
    for(int i=0;i<n;i++) scanf("%lld",&a[i]);
    printf("%lld\n",Solve(1,n-2,1,1)+a[0]+a[n-1]);
	return 0;
}
```