#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
inline int read()
{
	int x = 0 , f = 1; char ch = getchar();
	while(ch < '0' || ch > '9')	{if(ch == '-')	f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
const int N = 20;
ll a[N];
ll dfs(int l,int r,int xl,int xr)
{
	if(r - l <= 1) return 0;
	ll ans = 8e18;
	for(int i = l + 1;i < r;i ++) ans = min(ans,dfs(l,i,xl,xl + xr) + dfs(i,r,xl + xr,xr) + a[i] * (xl + xr));
	return ans;
}
int main()
{
	int n = read();
	for(int i = 1;i <= n;i ++) a[i] = read();
	printf("%lld\n",a[1] + a[n] + dfs(1,n,1,1));
}
### Prompt

Generate a CPP solution to 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 <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
inline int read()
{
	int x = 0 , f = 1; char ch = getchar();
	while(ch < '0' || ch > '9')	{if(ch == '-')	f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
const int N = 20;
ll a[N];
ll dfs(int l,int r,int xl,int xr)
{
	if(r - l <= 1) return 0;
	ll ans = 8e18;
	for(int i = l + 1;i < r;i ++) ans = min(ans,dfs(l,i,xl,xl + xr) + dfs(i,r,xl + xr,xr) + a[i] * (xl + xr));
	return ans;
}
int main()
{
	int n = read();
	for(int i = 1;i <= n;i ++) a[i] = read();
	printf("%lld\n",a[1] + a[n] + dfs(1,n,1,1));
}
```