#include <cstdio>
#include <map>
#include <vector>
using namespace std;


int n;
vector<long long> A;

map<pair<long long,long long>, long long> M;

const long long INF=1000000000000000000ll;

long long rec(int i,int j,long long a,long long b)
{
  if(i+1==j){
    return A[i]*a;
  }
  auto it=M.find(make_pair(i+n*a,j+n*b));
  if(it!=M.end()){
    return it->second;
  }
  long long ret=INF;
  for(int k=i+1;k<j;k++){
    ret=min(ret,rec(i,k,a,a+b)+rec(k,j,a+b,b));
  }
  M[make_pair(i+n*a,j+n*b)]=ret;
  return ret;
}

int main()
{
  scanf("%d",&n);
  A=vector<long long>(n);
  for(int i=0;i<n;i++){
    scanf("%lld",&A[i]);
  }
  printf("%lld\n",rec(0,n-1,1,1)+A[n-1]);
  return 0;
}

### Prompt

Create a solution in CPP for 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 <map>
#include <vector>
using namespace std;


int n;
vector<long long> A;

map<pair<long long,long long>, long long> M;

const long long INF=1000000000000000000ll;

long long rec(int i,int j,long long a,long long b)
{
  if(i+1==j){
    return A[i]*a;
  }
  auto it=M.find(make_pair(i+n*a,j+n*b));
  if(it!=M.end()){
    return it->second;
  }
  long long ret=INF;
  for(int k=i+1;k<j;k++){
    ret=min(ret,rec(i,k,a,a+b)+rec(k,j,a+b,b));
  }
  M[make_pair(i+n*a,j+n*b)]=ret;
  return ret;
}

int main()
{
  scanf("%d",&n);
  A=vector<long long>(n);
  for(int i=0;i<n;i++){
    scanf("%lld",&A[i]);
  }
  printf("%lld\n",rec(0,n-1,1,1)+A[n-1]);
  return 0;
}

```