#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pll pair<ll,ll>
#define inf 2333333333333333333LL

map<pll,ll> mp[20][20];

int num[110],n;

inline int rd()
{
	int x=0;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar());
	for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x;
}

inline ll gao(int l,int r,ll x,ll y)
{
	if (l>r) return 0;
	if (mp[l][r].count(pll(x,y))) return mp[l][r][pll(x,y)];
	ll res=inf;
	for (int i=l;i<=r;i++) res=min(res,num[i]*(x+y)+gao(l,i-1,x,x+y)+gao(i+1,r,x+y,y));
	return mp[l][r][pll(x,y)]=res;
}

int main()
{
	n=rd();
	for (int i=1;i<=n;i++) num[i]=rd();
	printf("%lld\n",gao(2,n-1,1,1)+num[1]+num[n]);
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 pll pair<ll,ll>
#define inf 2333333333333333333LL

map<pll,ll> mp[20][20];

int num[110],n;

inline int rd()
{
	int x=0;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar());
	for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x;
}

inline ll gao(int l,int r,ll x,ll y)
{
	if (l>r) return 0;
	if (mp[l][r].count(pll(x,y))) return mp[l][r][pll(x,y)];
	ll res=inf;
	for (int i=l;i<=r;i++) res=min(res,num[i]*(x+y)+gao(l,i-1,x,x+y)+gao(i+1,r,x+y,y));
	return mp[l][r][pll(x,y)]=res;
}

int main()
{
	n=rd();
	for (int i=1;i<=n;i++) num[i]=rd();
	printf("%lld\n",gao(2,n-1,1,1)+num[1]+num[n]);
	return 0;
}
```