#include<bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N=18;
const LL INF=(1LL<<60)-1;

int n;
LL a[N+9];

void into(){
  scanf("%d",&n);
  for (int i=1;i<=n;++i)
	scanf("%lld",&a[i]);
}

map<int,map<int,LL> >dp[N+9][N+9];

LL Dfs_dp(int l,int r,int x,int y){
  if (dp[l][r][x].find(y)!=dp[l][r][x].end()) return dp[l][r][x][y];
  if (l+1==r) return 0;
  LL res=INF;
  for (int i=l+1;i<r;++i) res=min(res,Dfs_dp(l,i,x,x+y)+Dfs_dp(i,r,x+y,y)+a[i]*(x+y));
  return dp[l][r][x][y]=res;
}

void outo(){
  printf("%lld\n",Dfs_dp(1,n,1,1)+a[1]+a[n]);
}

int main(){
  into();
  outo();
  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;

typedef long long LL;

const int N=18;
const LL INF=(1LL<<60)-1;

int n;
LL a[N+9];

void into(){
  scanf("%d",&n);
  for (int i=1;i<=n;++i)
	scanf("%lld",&a[i]);
}

map<int,map<int,LL> >dp[N+9][N+9];

LL Dfs_dp(int l,int r,int x,int y){
  if (dp[l][r][x].find(y)!=dp[l][r][x].end()) return dp[l][r][x][y];
  if (l+1==r) return 0;
  LL res=INF;
  for (int i=l+1;i<r;++i) res=min(res,Dfs_dp(l,i,x,x+y)+Dfs_dp(i,r,x+y,y)+a[i]*(x+y));
  return dp[l][r][x][y]=res;
}

void outo(){
  printf("%lld\n",Dfs_dp(1,n,1,1)+a[1]+a[n]);
}

int main(){
  into();
  outo();
  return 0;
}
```